Proving $(n+1)^2+(n+2)^2+…+(2n)^2= frac{n(2n+1)(7n+1)}{6}$ by induction












1














My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= frac{n(2n+1)(7n+1)}{6}$



My workings




  1. LHS=$2^2$ =$4$ RHS= $frac{24}{6} =4 $

  2. $(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= frac{k(2k+1)(7k+1)}{6}$

  3. LHS (subsituting $n= k+1$)----> $(k+2)^2+(k+3)^2+(k+4)^2+...+(2k)^2+(2k+1)^2$


Now, this is where my problem has started. When I substitute $n=k+1$ in the third step, I do not have the $(k+1)^2$ anymore. So I cannot use the statement in second step. So my question is what should i do next?






discrete-mathematics induction







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  • 1




    $2(k+1)=2k+2{}$
    – Lord Shark the Unknown
    Nov 24 at 11:48










  • @LordSharktheUnknown uh? sorry can you please provide some explanation please? Thank you.
    – Tfue
    Nov 24 at 11:50










  • @Tfue the last term in 3) must be $(2k+2)^2$ (not $(2k+1)^2$).
    – drhab
    Nov 24 at 11:53












  • @drhab why can you please give me some explantions? sorry and thank you.
    – Tfue
    Nov 24 at 11:57










  • If you substitute $n=k+1$ in $(n+1)^2+cdots+(2n)^2$ then you get $((k+1)+1)^2+cdots(2(k+1))^2=(k+2)^2+cdots+(2k+2)^2$.
    – drhab
    Nov 24 at 12:07
















1














My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= frac{n(2n+1)(7n+1)}{6}$



My workings




  1. LHS=$2^2$ =$4$ RHS= $frac{24}{6} =4 $

  2. $(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= frac{k(2k+1)(7k+1)}{6}$

  3. LHS (subsituting $n= k+1$)----> $(k+2)^2+(k+3)^2+(k+4)^2+...+(2k)^2+(2k+1)^2$


Now, this is where my problem has started. When I substitute $n=k+1$ in the third step, I do not have the $(k+1)^2$ anymore. So I cannot use the statement in second step. So my question is what should i do next?






discrete-mathematics induction







share|cite|improve this question




















  • 1




    $2(k+1)=2k+2{}$
    – Lord Shark the Unknown
    Nov 24 at 11:48










  • @LordSharktheUnknown uh? sorry can you please provide some explanation please? Thank you.
    – Tfue
    Nov 24 at 11:50










  • @Tfue the last term in 3) must be $(2k+2)^2$ (not $(2k+1)^2$).
    – drhab
    Nov 24 at 11:53












  • @drhab why can you please give me some explantions? sorry and thank you.
    – Tfue
    Nov 24 at 11:57










  • If you substitute $n=k+1$ in $(n+1)^2+cdots+(2n)^2$ then you get $((k+1)+1)^2+cdots(2(k+1))^2=(k+2)^2+cdots+(2k+2)^2$.
    – drhab
    Nov 24 at 12:07














1












1








1







My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= frac{n(2n+1)(7n+1)}{6}$



My workings




  1. LHS=$2^2$ =$4$ RHS= $frac{24}{6} =4 $

  2. $(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= frac{k(2k+1)(7k+1)}{6}$

  3. LHS (subsituting $n= k+1$)----> $(k+2)^2+(k+3)^2+(k+4)^2+...+(2k)^2+(2k+1)^2$


Now, this is where my problem has started. When I substitute $n=k+1$ in the third step, I do not have the $(k+1)^2$ anymore. So I cannot use the statement in second step. So my question is what should i do next?






discrete-mathematics induction







share|cite|improve this question















My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= frac{n(2n+1)(7n+1)}{6}$



My workings




  1. LHS=$2^2$ =$4$ RHS= $frac{24}{6} =4 $

  2. $(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= frac{k(2k+1)(7k+1)}{6}$

  3. LHS (subsituting $n= k+1$)----> $(k+2)^2+(k+3)^2+(k+4)^2+...+(2k)^2+(2k+1)^2$


Now, this is where my problem has started. When I substitute $n=k+1$ in the third step, I do not have the $(k+1)^2$ anymore. So I cannot use the statement in second step. So my question is what should i do next?






discrete-mathematics induction




discrete-mathematics induction






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edited Nov 24 at 13:39









user21820

38.6k542153




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asked Nov 24 at 11:38









Tfue

859




859








  • 1




    $2(k+1)=2k+2{}$
    – Lord Shark the Unknown
    Nov 24 at 11:48










  • @LordSharktheUnknown uh? sorry can you please provide some explanation please? Thank you.
    – Tfue
    Nov 24 at 11:50










  • @Tfue the last term in 3) must be $(2k+2)^2$ (not $(2k+1)^2$).
    – drhab
    Nov 24 at 11:53












  • @drhab why can you please give me some explantions? sorry and thank you.
    – Tfue
    Nov 24 at 11:57










  • If you substitute $n=k+1$ in $(n+1)^2+cdots+(2n)^2$ then you get $((k+1)+1)^2+cdots(2(k+1))^2=(k+2)^2+cdots+(2k+2)^2$.
    – drhab
    Nov 24 at 12:07














  • 1




    $2(k+1)=2k+2{}$
    – Lord Shark the Unknown
    Nov 24 at 11:48










  • @LordSharktheUnknown uh? sorry can you please provide some explanation please? Thank you.
    – Tfue
    Nov 24 at 11:50










  • @Tfue the last term in 3) must be $(2k+2)^2$ (not $(2k+1)^2$).
    – drhab
    Nov 24 at 11:53












  • @drhab why can you please give me some explantions? sorry and thank you.
    – Tfue
    Nov 24 at 11:57










  • If you substitute $n=k+1$ in $(n+1)^2+cdots+(2n)^2$ then you get $((k+1)+1)^2+cdots(2(k+1))^2=(k+2)^2+cdots+(2k+2)^2$.
    – drhab
    Nov 24 at 12:07








1




1




$2(k+1)=2k+2{}$
– Lord Shark the Unknown
Nov 24 at 11:48




$2(k+1)=2k+2{}$
– Lord Shark the Unknown
Nov 24 at 11:48












@LordSharktheUnknown uh? sorry can you please provide some explanation please? Thank you.
– Tfue
Nov 24 at 11:50




@LordSharktheUnknown uh? sorry can you please provide some explanation please? Thank you.
– Tfue
Nov 24 at 11:50












@Tfue the last term in 3) must be $(2k+2)^2$ (not $(2k+1)^2$).
– drhab
Nov 24 at 11:53






@Tfue the last term in 3) must be $(2k+2)^2$ (not $(2k+1)^2$).
– drhab
Nov 24 at 11:53














@drhab why can you please give me some explantions? sorry and thank you.
– Tfue
Nov 24 at 11:57




@drhab why can you please give me some explantions? sorry and thank you.
– Tfue
Nov 24 at 11:57












If you substitute $n=k+1$ in $(n+1)^2+cdots+(2n)^2$ then you get $((k+1)+1)^2+cdots(2(k+1))^2=(k+2)^2+cdots+(2k+2)^2$.
– drhab
Nov 24 at 12:07




If you substitute $n=k+1$ in $(n+1)^2+cdots+(2n)^2$ then you get $((k+1)+1)^2+cdots(2(k+1))^2=(k+2)^2+cdots+(2k+2)^2$.
– drhab
Nov 24 at 12:07










5 Answers
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2














This is 2nd tough-degree exercise in induction. Observe that both the first and last elements change, so if $;k=n;$ we get



$$(n+1)^2+(n+2)^2+ldots+(2n)^2=frac{n(2n+1)(7n+1)}6;;(**)$$



then for $;k=n+1;$ we get



$$(n+1+1)^2+(n+3)^2+ldots+(2n)^2+(2n+1)^2+overbrace{(2n+2)^2}^{=(2(n+1))^2}=frac{(n+1)(2n+3)(7n+8)}6$$



Now, we prove the last one assuming the first one. Developing left side:



$$(n+2)^2+ldots+(2n)^2+(2n+1)^2+(2n+2)^2=(n+1)^2+ldots+(2n+2)^2-(n+1)^2=$$



$$=(n+1)^2+ldots+(2n)^2+(2n+1)^2+(2n+2)^2-(n+1)^2stackrel{text{Ind. Hyp.}}=$$



$$=frac{n(2n+1)(7n+1)}6+(2n+1)^2+4(n+1)^2-(n+1)^2=$$



$$=frac{n(2n+1)(7n+1)}6+(2n+1)^2+3(n+1)^2=frac{(2n+1)left[n(7n+1)+12n+6right]+18(n+1)^2}6=$$



$$=frac{(2n+1)(7n^2+13n+6)+18n^2+36n+18}6=frac{(n+1)(2n+3)(7n+8)}6$$



The last equality you can prove it opening parentheses and equating coefficients, or showing the left side has as roots $;-1,,-3/2,,-8/7;$



Alternative way to prove:



$$(n+1)^2+(n+2)^2+ldots+(n+n)^2=$$



$$=n^2+ldots+n^2+2n+4n+ldots+2nn+1+2^2+ldots+n^2=$$



$$ncdot n^2+2(1+2+ldots+n)n+frac{n(n+1)(2n+1)}6=$$



$$=n^3+n^2(n+1)+frac{n(n+1)(2n+1)}6=ldots$$






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  • Wow thanks for this detailed solution! However i still have a every beginnernish question, why did you add $(2n+2)^2$?(in the first line for $k=n+1$ Sorry and thank you.
    – Tfue
    Nov 24 at 12:04












  • Because the rule is: "Begin with some natural number $;n;$ when one is added, and end with twice that number $;n;$ "... Thus, when in the inductive step with begin with $;n+1;$ and add one to it, we must end with $;2(n+1);$ ...But we sum consecutive naturals, so if the last one in the first step is $;(2n)^2;$ , the last one in the ind. step is $;(2(n+1))^2;$ , and there are two consecutive integers there: $;(2n)^2+color{red}{(2n+1)^2}+color{red}{(2n+2)^2};$ ...This last addend is just $;2n+2=2(n+1);$ ...the first one times two! (and all squared, of course)
    – DonAntonio
    Nov 24 at 12:11












  • Sorry sir i meant to write how did you obtain $(2n+1)^2$ which this i accidentally obtained. I do get the $(2n+2)^2$ part now thank you.
    – Tfue
    Nov 24 at 12:22










  • sorry sir i could not understand what you meant.
    – Tfue
    Nov 25 at 6:52



















1














Your third step is wrong. What is to be proven is the following on substituting n = k+1,$$(k+2)^2+(k+3)^2+...+(2k+2)^2 = frac{(k+1)(2k+3)(7k+8)}{6}$$.
We know the result for $(k+1)^2+(k+2)^2+...+(2k)^2$, substitute that here, you'll get,$$LHS = frac{(k)(2k+1)(7k+1)}{6} + (2k+1)^2+(2k+2)^2-(k+1)^2$$
$$ = frac{(k+1)(2k+3)(7k+8)}{6}$$



Hope it helps






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  • Hey thanks for pointing out mistakes and presenting a solution! However i still have a question. Seeing that i accidentally obtained $(2k+1)^2$ which i meant to write $(2k+2)^2$. Can you please explain ton me how did you get $(2k+1)^2$ and $-(k+1)^2$
    – Tfue
    Nov 24 at 12:20










  • This is how we use induction, we use the induction hypothesis in the induction step for the terms which are common and solve for the rest to get the desired result.
    – Martund
    Nov 24 at 12:43



















1














Assuming the step for $n=k$ $:$For



$n=k+1$ $(k+2)^2+cdots+(2k+2)^2=dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2$ $=dfrac{k(2k+1)(7k+1)}{6}+3(k+1)^2+(2k+1)^2=dfrac{k(2k+1)(7k+1)}{6}+7k^2+10k+4=dfrac{(k+1)(2k+3)(7k+8)}{6}$






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  • Thanks for answering! However i still have a question how did you get $-(k+1)^2$ and the $(2k+1)^2$
    – Tfue
    Nov 24 at 12:17










  • $(k+2)^2+cdots+(2k)^2=dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2-(k+1)^2=(k+2)^2cdots+(2k)^2$ which you can see from $n=k$
    – Yadati Kiran
    Nov 24 at 12:37












  • When you substitute $n=k+1$ in $(2k)^2$ we get $(2k+2)^2$. Hence the sum for $n=k+1$ is $(k+2)^2+cdots+(2k)^2+(2k+1)^2+(2k+2)^2$.
    – Yadati Kiran
    Nov 24 at 12:38










  • Can't i just do $(k+2)^2+...+(2k)^2+(2k+2)^2$
    – Tfue
    Nov 24 at 12:41










  • No. Just put some values for $k$ to get clarity
    – Yadati Kiran
    Nov 24 at 12:42



















0














Observe that:



$$(k+2)^2+cdots+(2k+2)^2=$$$$left[(k+1)^2+(k+2)^2+cdots+(2k)^2right]+(2k+1)^2+(2k+2)^2-(k+1)^2=$$$$f(k)+(2k+1)^2+(2k+2)^2-(k+1)^2$$



where $f(k)$ denotes the RHS of 2).



If the equality indeed holds then the outcome must be $f(k+1)$ and that can be checked.






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    0














    $(k+2)^2+dots+(2k)^2+(2k+1)^2+(2k+2)^2=frac{k(2k+1)(7k+1)}6-(k+1)^2+(2k+1)^2+(2k+2)^2=frac{k(14k^2+9k+1)+6(7k^2+10k+4)}6=frac{(k+1)(2k+3)(7k+8)}6$.






    share|cite|improve this answer





















    • Hey thanks can you please explain how did you get $(2k+1)^2$ thank you.
      – Tfue
      Nov 25 at 7:03










    • You need all of the $(k+2)^2$ through $(2k+2)^2$.
      – Chris Custer
      Nov 25 at 7:49











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    5 Answers
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    5 Answers
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    active

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    active

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    active

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    2














    This is 2nd tough-degree exercise in induction. Observe that both the first and last elements change, so if $;k=n;$ we get



    $$(n+1)^2+(n+2)^2+ldots+(2n)^2=frac{n(2n+1)(7n+1)}6;;(**)$$



    then for $;k=n+1;$ we get



    $$(n+1+1)^2+(n+3)^2+ldots+(2n)^2+(2n+1)^2+overbrace{(2n+2)^2}^{=(2(n+1))^2}=frac{(n+1)(2n+3)(7n+8)}6$$



    Now, we prove the last one assuming the first one. Developing left side:



    $$(n+2)^2+ldots+(2n)^2+(2n+1)^2+(2n+2)^2=(n+1)^2+ldots+(2n+2)^2-(n+1)^2=$$



    $$=(n+1)^2+ldots+(2n)^2+(2n+1)^2+(2n+2)^2-(n+1)^2stackrel{text{Ind. Hyp.}}=$$



    $$=frac{n(2n+1)(7n+1)}6+(2n+1)^2+4(n+1)^2-(n+1)^2=$$



    $$=frac{n(2n+1)(7n+1)}6+(2n+1)^2+3(n+1)^2=frac{(2n+1)left[n(7n+1)+12n+6right]+18(n+1)^2}6=$$



    $$=frac{(2n+1)(7n^2+13n+6)+18n^2+36n+18}6=frac{(n+1)(2n+3)(7n+8)}6$$



    The last equality you can prove it opening parentheses and equating coefficients, or showing the left side has as roots $;-1,,-3/2,,-8/7;$



    Alternative way to prove:



    $$(n+1)^2+(n+2)^2+ldots+(n+n)^2=$$



    $$=n^2+ldots+n^2+2n+4n+ldots+2nn+1+2^2+ldots+n^2=$$



    $$ncdot n^2+2(1+2+ldots+n)n+frac{n(n+1)(2n+1)}6=$$



    $$=n^3+n^2(n+1)+frac{n(n+1)(2n+1)}6=ldots$$






    share|cite|improve this answer























    • Wow thanks for this detailed solution! However i still have a every beginnernish question, why did you add $(2n+2)^2$?(in the first line for $k=n+1$ Sorry and thank you.
      – Tfue
      Nov 24 at 12:04












    • Because the rule is: "Begin with some natural number $;n;$ when one is added, and end with twice that number $;n;$ "... Thus, when in the inductive step with begin with $;n+1;$ and add one to it, we must end with $;2(n+1);$ ...But we sum consecutive naturals, so if the last one in the first step is $;(2n)^2;$ , the last one in the ind. step is $;(2(n+1))^2;$ , and there are two consecutive integers there: $;(2n)^2+color{red}{(2n+1)^2}+color{red}{(2n+2)^2};$ ...This last addend is just $;2n+2=2(n+1);$ ...the first one times two! (and all squared, of course)
      – DonAntonio
      Nov 24 at 12:11












    • Sorry sir i meant to write how did you obtain $(2n+1)^2$ which this i accidentally obtained. I do get the $(2n+2)^2$ part now thank you.
      – Tfue
      Nov 24 at 12:22










    • sorry sir i could not understand what you meant.
      – Tfue
      Nov 25 at 6:52
















    2














    This is 2nd tough-degree exercise in induction. Observe that both the first and last elements change, so if $;k=n;$ we get



    $$(n+1)^2+(n+2)^2+ldots+(2n)^2=frac{n(2n+1)(7n+1)}6;;(**)$$



    then for $;k=n+1;$ we get



    $$(n+1+1)^2+(n+3)^2+ldots+(2n)^2+(2n+1)^2+overbrace{(2n+2)^2}^{=(2(n+1))^2}=frac{(n+1)(2n+3)(7n+8)}6$$



    Now, we prove the last one assuming the first one. Developing left side:



    $$(n+2)^2+ldots+(2n)^2+(2n+1)^2+(2n+2)^2=(n+1)^2+ldots+(2n+2)^2-(n+1)^2=$$



    $$=(n+1)^2+ldots+(2n)^2+(2n+1)^2+(2n+2)^2-(n+1)^2stackrel{text{Ind. Hyp.}}=$$



    $$=frac{n(2n+1)(7n+1)}6+(2n+1)^2+4(n+1)^2-(n+1)^2=$$



    $$=frac{n(2n+1)(7n+1)}6+(2n+1)^2+3(n+1)^2=frac{(2n+1)left[n(7n+1)+12n+6right]+18(n+1)^2}6=$$



    $$=frac{(2n+1)(7n^2+13n+6)+18n^2+36n+18}6=frac{(n+1)(2n+3)(7n+8)}6$$



    The last equality you can prove it opening parentheses and equating coefficients, or showing the left side has as roots $;-1,,-3/2,,-8/7;$



    Alternative way to prove:



    $$(n+1)^2+(n+2)^2+ldots+(n+n)^2=$$



    $$=n^2+ldots+n^2+2n+4n+ldots+2nn+1+2^2+ldots+n^2=$$



    $$ncdot n^2+2(1+2+ldots+n)n+frac{n(n+1)(2n+1)}6=$$



    $$=n^3+n^2(n+1)+frac{n(n+1)(2n+1)}6=ldots$$






    share|cite|improve this answer























    • Wow thanks for this detailed solution! However i still have a every beginnernish question, why did you add $(2n+2)^2$?(in the first line for $k=n+1$ Sorry and thank you.
      – Tfue
      Nov 24 at 12:04












    • Because the rule is: "Begin with some natural number $;n;$ when one is added, and end with twice that number $;n;$ "... Thus, when in the inductive step with begin with $;n+1;$ and add one to it, we must end with $;2(n+1);$ ...But we sum consecutive naturals, so if the last one in the first step is $;(2n)^2;$ , the last one in the ind. step is $;(2(n+1))^2;$ , and there are two consecutive integers there: $;(2n)^2+color{red}{(2n+1)^2}+color{red}{(2n+2)^2};$ ...This last addend is just $;2n+2=2(n+1);$ ...the first one times two! (and all squared, of course)
      – DonAntonio
      Nov 24 at 12:11












    • Sorry sir i meant to write how did you obtain $(2n+1)^2$ which this i accidentally obtained. I do get the $(2n+2)^2$ part now thank you.
      – Tfue
      Nov 24 at 12:22










    • sorry sir i could not understand what you meant.
      – Tfue
      Nov 25 at 6:52














    2












    2








    2






    This is 2nd tough-degree exercise in induction. Observe that both the first and last elements change, so if $;k=n;$ we get



    $$(n+1)^2+(n+2)^2+ldots+(2n)^2=frac{n(2n+1)(7n+1)}6;;(**)$$



    then for $;k=n+1;$ we get



    $$(n+1+1)^2+(n+3)^2+ldots+(2n)^2+(2n+1)^2+overbrace{(2n+2)^2}^{=(2(n+1))^2}=frac{(n+1)(2n+3)(7n+8)}6$$



    Now, we prove the last one assuming the first one. Developing left side:



    $$(n+2)^2+ldots+(2n)^2+(2n+1)^2+(2n+2)^2=(n+1)^2+ldots+(2n+2)^2-(n+1)^2=$$



    $$=(n+1)^2+ldots+(2n)^2+(2n+1)^2+(2n+2)^2-(n+1)^2stackrel{text{Ind. Hyp.}}=$$



    $$=frac{n(2n+1)(7n+1)}6+(2n+1)^2+4(n+1)^2-(n+1)^2=$$



    $$=frac{n(2n+1)(7n+1)}6+(2n+1)^2+3(n+1)^2=frac{(2n+1)left[n(7n+1)+12n+6right]+18(n+1)^2}6=$$



    $$=frac{(2n+1)(7n^2+13n+6)+18n^2+36n+18}6=frac{(n+1)(2n+3)(7n+8)}6$$



    The last equality you can prove it opening parentheses and equating coefficients, or showing the left side has as roots $;-1,,-3/2,,-8/7;$



    Alternative way to prove:



    $$(n+1)^2+(n+2)^2+ldots+(n+n)^2=$$



    $$=n^2+ldots+n^2+2n+4n+ldots+2nn+1+2^2+ldots+n^2=$$



    $$ncdot n^2+2(1+2+ldots+n)n+frac{n(n+1)(2n+1)}6=$$



    $$=n^3+n^2(n+1)+frac{n(n+1)(2n+1)}6=ldots$$






    share|cite|improve this answer














    This is 2nd tough-degree exercise in induction. Observe that both the first and last elements change, so if $;k=n;$ we get



    $$(n+1)^2+(n+2)^2+ldots+(2n)^2=frac{n(2n+1)(7n+1)}6;;(**)$$



    then for $;k=n+1;$ we get



    $$(n+1+1)^2+(n+3)^2+ldots+(2n)^2+(2n+1)^2+overbrace{(2n+2)^2}^{=(2(n+1))^2}=frac{(n+1)(2n+3)(7n+8)}6$$



    Now, we prove the last one assuming the first one. Developing left side:



    $$(n+2)^2+ldots+(2n)^2+(2n+1)^2+(2n+2)^2=(n+1)^2+ldots+(2n+2)^2-(n+1)^2=$$



    $$=(n+1)^2+ldots+(2n)^2+(2n+1)^2+(2n+2)^2-(n+1)^2stackrel{text{Ind. Hyp.}}=$$



    $$=frac{n(2n+1)(7n+1)}6+(2n+1)^2+4(n+1)^2-(n+1)^2=$$



    $$=frac{n(2n+1)(7n+1)}6+(2n+1)^2+3(n+1)^2=frac{(2n+1)left[n(7n+1)+12n+6right]+18(n+1)^2}6=$$



    $$=frac{(2n+1)(7n^2+13n+6)+18n^2+36n+18}6=frac{(n+1)(2n+3)(7n+8)}6$$



    The last equality you can prove it opening parentheses and equating coefficients, or showing the left side has as roots $;-1,,-3/2,,-8/7;$



    Alternative way to prove:



    $$(n+1)^2+(n+2)^2+ldots+(n+n)^2=$$



    $$=n^2+ldots+n^2+2n+4n+ldots+2nn+1+2^2+ldots+n^2=$$



    $$ncdot n^2+2(1+2+ldots+n)n+frac{n(n+1)(2n+1)}6=$$



    $$=n^3+n^2(n+1)+frac{n(n+1)(2n+1)}6=ldots$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 24 at 12:08

























    answered Nov 24 at 11:58









    DonAntonio

    176k1491225




    176k1491225












    • Wow thanks for this detailed solution! However i still have a every beginnernish question, why did you add $(2n+2)^2$?(in the first line for $k=n+1$ Sorry and thank you.
      – Tfue
      Nov 24 at 12:04












    • Because the rule is: "Begin with some natural number $;n;$ when one is added, and end with twice that number $;n;$ "... Thus, when in the inductive step with begin with $;n+1;$ and add one to it, we must end with $;2(n+1);$ ...But we sum consecutive naturals, so if the last one in the first step is $;(2n)^2;$ , the last one in the ind. step is $;(2(n+1))^2;$ , and there are two consecutive integers there: $;(2n)^2+color{red}{(2n+1)^2}+color{red}{(2n+2)^2};$ ...This last addend is just $;2n+2=2(n+1);$ ...the first one times two! (and all squared, of course)
      – DonAntonio
      Nov 24 at 12:11












    • Sorry sir i meant to write how did you obtain $(2n+1)^2$ which this i accidentally obtained. I do get the $(2n+2)^2$ part now thank you.
      – Tfue
      Nov 24 at 12:22










    • sorry sir i could not understand what you meant.
      – Tfue
      Nov 25 at 6:52


















    • Wow thanks for this detailed solution! However i still have a every beginnernish question, why did you add $(2n+2)^2$?(in the first line for $k=n+1$ Sorry and thank you.
      – Tfue
      Nov 24 at 12:04












    • Because the rule is: "Begin with some natural number $;n;$ when one is added, and end with twice that number $;n;$ "... Thus, when in the inductive step with begin with $;n+1;$ and add one to it, we must end with $;2(n+1);$ ...But we sum consecutive naturals, so if the last one in the first step is $;(2n)^2;$ , the last one in the ind. step is $;(2(n+1))^2;$ , and there are two consecutive integers there: $;(2n)^2+color{red}{(2n+1)^2}+color{red}{(2n+2)^2};$ ...This last addend is just $;2n+2=2(n+1);$ ...the first one times two! (and all squared, of course)
      – DonAntonio
      Nov 24 at 12:11












    • Sorry sir i meant to write how did you obtain $(2n+1)^2$ which this i accidentally obtained. I do get the $(2n+2)^2$ part now thank you.
      – Tfue
      Nov 24 at 12:22










    • sorry sir i could not understand what you meant.
      – Tfue
      Nov 25 at 6:52
















    Wow thanks for this detailed solution! However i still have a every beginnernish question, why did you add $(2n+2)^2$?(in the first line for $k=n+1$ Sorry and thank you.
    – Tfue
    Nov 24 at 12:04






    Wow thanks for this detailed solution! However i still have a every beginnernish question, why did you add $(2n+2)^2$?(in the first line for $k=n+1$ Sorry and thank you.
    – Tfue
    Nov 24 at 12:04














    Because the rule is: "Begin with some natural number $;n;$ when one is added, and end with twice that number $;n;$ "... Thus, when in the inductive step with begin with $;n+1;$ and add one to it, we must end with $;2(n+1);$ ...But we sum consecutive naturals, so if the last one in the first step is $;(2n)^2;$ , the last one in the ind. step is $;(2(n+1))^2;$ , and there are two consecutive integers there: $;(2n)^2+color{red}{(2n+1)^2}+color{red}{(2n+2)^2};$ ...This last addend is just $;2n+2=2(n+1);$ ...the first one times two! (and all squared, of course)
    – DonAntonio
    Nov 24 at 12:11






    Because the rule is: "Begin with some natural number $;n;$ when one is added, and end with twice that number $;n;$ "... Thus, when in the inductive step with begin with $;n+1;$ and add one to it, we must end with $;2(n+1);$ ...But we sum consecutive naturals, so if the last one in the first step is $;(2n)^2;$ , the last one in the ind. step is $;(2(n+1))^2;$ , and there are two consecutive integers there: $;(2n)^2+color{red}{(2n+1)^2}+color{red}{(2n+2)^2};$ ...This last addend is just $;2n+2=2(n+1);$ ...the first one times two! (and all squared, of course)
    – DonAntonio
    Nov 24 at 12:11














    Sorry sir i meant to write how did you obtain $(2n+1)^2$ which this i accidentally obtained. I do get the $(2n+2)^2$ part now thank you.
    – Tfue
    Nov 24 at 12:22




    Sorry sir i meant to write how did you obtain $(2n+1)^2$ which this i accidentally obtained. I do get the $(2n+2)^2$ part now thank you.
    – Tfue
    Nov 24 at 12:22












    sorry sir i could not understand what you meant.
    – Tfue
    Nov 25 at 6:52




    sorry sir i could not understand what you meant.
    – Tfue
    Nov 25 at 6:52











    1














    Your third step is wrong. What is to be proven is the following on substituting n = k+1,$$(k+2)^2+(k+3)^2+...+(2k+2)^2 = frac{(k+1)(2k+3)(7k+8)}{6}$$.
    We know the result for $(k+1)^2+(k+2)^2+...+(2k)^2$, substitute that here, you'll get,$$LHS = frac{(k)(2k+1)(7k+1)}{6} + (2k+1)^2+(2k+2)^2-(k+1)^2$$
    $$ = frac{(k+1)(2k+3)(7k+8)}{6}$$



    Hope it helps






    share|cite|improve this answer





















    • Hey thanks for pointing out mistakes and presenting a solution! However i still have a question. Seeing that i accidentally obtained $(2k+1)^2$ which i meant to write $(2k+2)^2$. Can you please explain ton me how did you get $(2k+1)^2$ and $-(k+1)^2$
      – Tfue
      Nov 24 at 12:20










    • This is how we use induction, we use the induction hypothesis in the induction step for the terms which are common and solve for the rest to get the desired result.
      – Martund
      Nov 24 at 12:43
















    1














    Your third step is wrong. What is to be proven is the following on substituting n = k+1,$$(k+2)^2+(k+3)^2+...+(2k+2)^2 = frac{(k+1)(2k+3)(7k+8)}{6}$$.
    We know the result for $(k+1)^2+(k+2)^2+...+(2k)^2$, substitute that here, you'll get,$$LHS = frac{(k)(2k+1)(7k+1)}{6} + (2k+1)^2+(2k+2)^2-(k+1)^2$$
    $$ = frac{(k+1)(2k+3)(7k+8)}{6}$$



    Hope it helps






    share|cite|improve this answer





















    • Hey thanks for pointing out mistakes and presenting a solution! However i still have a question. Seeing that i accidentally obtained $(2k+1)^2$ which i meant to write $(2k+2)^2$. Can you please explain ton me how did you get $(2k+1)^2$ and $-(k+1)^2$
      – Tfue
      Nov 24 at 12:20










    • This is how we use induction, we use the induction hypothesis in the induction step for the terms which are common and solve for the rest to get the desired result.
      – Martund
      Nov 24 at 12:43














    1












    1








    1






    Your third step is wrong. What is to be proven is the following on substituting n = k+1,$$(k+2)^2+(k+3)^2+...+(2k+2)^2 = frac{(k+1)(2k+3)(7k+8)}{6}$$.
    We know the result for $(k+1)^2+(k+2)^2+...+(2k)^2$, substitute that here, you'll get,$$LHS = frac{(k)(2k+1)(7k+1)}{6} + (2k+1)^2+(2k+2)^2-(k+1)^2$$
    $$ = frac{(k+1)(2k+3)(7k+8)}{6}$$



    Hope it helps






    share|cite|improve this answer












    Your third step is wrong. What is to be proven is the following on substituting n = k+1,$$(k+2)^2+(k+3)^2+...+(2k+2)^2 = frac{(k+1)(2k+3)(7k+8)}{6}$$.
    We know the result for $(k+1)^2+(k+2)^2+...+(2k)^2$, substitute that here, you'll get,$$LHS = frac{(k)(2k+1)(7k+1)}{6} + (2k+1)^2+(2k+2)^2-(k+1)^2$$
    $$ = frac{(k+1)(2k+3)(7k+8)}{6}$$



    Hope it helps







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 24 at 11:53









    Martund

    1,344212




    1,344212












    • Hey thanks for pointing out mistakes and presenting a solution! However i still have a question. Seeing that i accidentally obtained $(2k+1)^2$ which i meant to write $(2k+2)^2$. Can you please explain ton me how did you get $(2k+1)^2$ and $-(k+1)^2$
      – Tfue
      Nov 24 at 12:20










    • This is how we use induction, we use the induction hypothesis in the induction step for the terms which are common and solve for the rest to get the desired result.
      – Martund
      Nov 24 at 12:43


















    • Hey thanks for pointing out mistakes and presenting a solution! However i still have a question. Seeing that i accidentally obtained $(2k+1)^2$ which i meant to write $(2k+2)^2$. Can you please explain ton me how did you get $(2k+1)^2$ and $-(k+1)^2$
      – Tfue
      Nov 24 at 12:20










    • This is how we use induction, we use the induction hypothesis in the induction step for the terms which are common and solve for the rest to get the desired result.
      – Martund
      Nov 24 at 12:43
















    Hey thanks for pointing out mistakes and presenting a solution! However i still have a question. Seeing that i accidentally obtained $(2k+1)^2$ which i meant to write $(2k+2)^2$. Can you please explain ton me how did you get $(2k+1)^2$ and $-(k+1)^2$
    – Tfue
    Nov 24 at 12:20




    Hey thanks for pointing out mistakes and presenting a solution! However i still have a question. Seeing that i accidentally obtained $(2k+1)^2$ which i meant to write $(2k+2)^2$. Can you please explain ton me how did you get $(2k+1)^2$ and $-(k+1)^2$
    – Tfue
    Nov 24 at 12:20












    This is how we use induction, we use the induction hypothesis in the induction step for the terms which are common and solve for the rest to get the desired result.
    – Martund
    Nov 24 at 12:43




    This is how we use induction, we use the induction hypothesis in the induction step for the terms which are common and solve for the rest to get the desired result.
    – Martund
    Nov 24 at 12:43











    1














    Assuming the step for $n=k$ $:$For



    $n=k+1$ $(k+2)^2+cdots+(2k+2)^2=dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2$ $=dfrac{k(2k+1)(7k+1)}{6}+3(k+1)^2+(2k+1)^2=dfrac{k(2k+1)(7k+1)}{6}+7k^2+10k+4=dfrac{(k+1)(2k+3)(7k+8)}{6}$






    share|cite|improve this answer





















    • Thanks for answering! However i still have a question how did you get $-(k+1)^2$ and the $(2k+1)^2$
      – Tfue
      Nov 24 at 12:17










    • $(k+2)^2+cdots+(2k)^2=dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2-(k+1)^2=(k+2)^2cdots+(2k)^2$ which you can see from $n=k$
      – Yadati Kiran
      Nov 24 at 12:37












    • When you substitute $n=k+1$ in $(2k)^2$ we get $(2k+2)^2$. Hence the sum for $n=k+1$ is $(k+2)^2+cdots+(2k)^2+(2k+1)^2+(2k+2)^2$.
      – Yadati Kiran
      Nov 24 at 12:38










    • Can't i just do $(k+2)^2+...+(2k)^2+(2k+2)^2$
      – Tfue
      Nov 24 at 12:41










    • No. Just put some values for $k$ to get clarity
      – Yadati Kiran
      Nov 24 at 12:42
















    1














    Assuming the step for $n=k$ $:$For



    $n=k+1$ $(k+2)^2+cdots+(2k+2)^2=dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2$ $=dfrac{k(2k+1)(7k+1)}{6}+3(k+1)^2+(2k+1)^2=dfrac{k(2k+1)(7k+1)}{6}+7k^2+10k+4=dfrac{(k+1)(2k+3)(7k+8)}{6}$






    share|cite|improve this answer





















    • Thanks for answering! However i still have a question how did you get $-(k+1)^2$ and the $(2k+1)^2$
      – Tfue
      Nov 24 at 12:17










    • $(k+2)^2+cdots+(2k)^2=dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2-(k+1)^2=(k+2)^2cdots+(2k)^2$ which you can see from $n=k$
      – Yadati Kiran
      Nov 24 at 12:37












    • When you substitute $n=k+1$ in $(2k)^2$ we get $(2k+2)^2$. Hence the sum for $n=k+1$ is $(k+2)^2+cdots+(2k)^2+(2k+1)^2+(2k+2)^2$.
      – Yadati Kiran
      Nov 24 at 12:38










    • Can't i just do $(k+2)^2+...+(2k)^2+(2k+2)^2$
      – Tfue
      Nov 24 at 12:41










    • No. Just put some values for $k$ to get clarity
      – Yadati Kiran
      Nov 24 at 12:42














    1












    1








    1






    Assuming the step for $n=k$ $:$For



    $n=k+1$ $(k+2)^2+cdots+(2k+2)^2=dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2$ $=dfrac{k(2k+1)(7k+1)}{6}+3(k+1)^2+(2k+1)^2=dfrac{k(2k+1)(7k+1)}{6}+7k^2+10k+4=dfrac{(k+1)(2k+3)(7k+8)}{6}$






    share|cite|improve this answer












    Assuming the step for $n=k$ $:$For



    $n=k+1$ $(k+2)^2+cdots+(2k+2)^2=dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2$ $=dfrac{k(2k+1)(7k+1)}{6}+3(k+1)^2+(2k+1)^2=dfrac{k(2k+1)(7k+1)}{6}+7k^2+10k+4=dfrac{(k+1)(2k+3)(7k+8)}{6}$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 24 at 11:54









    Yadati Kiran

    1,688519




    1,688519












    • Thanks for answering! However i still have a question how did you get $-(k+1)^2$ and the $(2k+1)^2$
      – Tfue
      Nov 24 at 12:17










    • $(k+2)^2+cdots+(2k)^2=dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2-(k+1)^2=(k+2)^2cdots+(2k)^2$ which you can see from $n=k$
      – Yadati Kiran
      Nov 24 at 12:37












    • When you substitute $n=k+1$ in $(2k)^2$ we get $(2k+2)^2$. Hence the sum for $n=k+1$ is $(k+2)^2+cdots+(2k)^2+(2k+1)^2+(2k+2)^2$.
      – Yadati Kiran
      Nov 24 at 12:38










    • Can't i just do $(k+2)^2+...+(2k)^2+(2k+2)^2$
      – Tfue
      Nov 24 at 12:41










    • No. Just put some values for $k$ to get clarity
      – Yadati Kiran
      Nov 24 at 12:42


















    • Thanks for answering! However i still have a question how did you get $-(k+1)^2$ and the $(2k+1)^2$
      – Tfue
      Nov 24 at 12:17










    • $(k+2)^2+cdots+(2k)^2=dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2-(k+1)^2=(k+2)^2cdots+(2k)^2$ which you can see from $n=k$
      – Yadati Kiran
      Nov 24 at 12:37












    • When you substitute $n=k+1$ in $(2k)^2$ we get $(2k+2)^2$. Hence the sum for $n=k+1$ is $(k+2)^2+cdots+(2k)^2+(2k+1)^2+(2k+2)^2$.
      – Yadati Kiran
      Nov 24 at 12:38










    • Can't i just do $(k+2)^2+...+(2k)^2+(2k+2)^2$
      – Tfue
      Nov 24 at 12:41










    • No. Just put some values for $k$ to get clarity
      – Yadati Kiran
      Nov 24 at 12:42
















    Thanks for answering! However i still have a question how did you get $-(k+1)^2$ and the $(2k+1)^2$
    – Tfue
    Nov 24 at 12:17




    Thanks for answering! However i still have a question how did you get $-(k+1)^2$ and the $(2k+1)^2$
    – Tfue
    Nov 24 at 12:17












    $(k+2)^2+cdots+(2k)^2=dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2-(k+1)^2=(k+2)^2cdots+(2k)^2$ which you can see from $n=k$
    – Yadati Kiran
    Nov 24 at 12:37






    $(k+2)^2+cdots+(2k)^2=dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2-(k+1)^2=(k+2)^2cdots+(2k)^2$ which you can see from $n=k$
    – Yadati Kiran
    Nov 24 at 12:37














    When you substitute $n=k+1$ in $(2k)^2$ we get $(2k+2)^2$. Hence the sum for $n=k+1$ is $(k+2)^2+cdots+(2k)^2+(2k+1)^2+(2k+2)^2$.
    – Yadati Kiran
    Nov 24 at 12:38




    When you substitute $n=k+1$ in $(2k)^2$ we get $(2k+2)^2$. Hence the sum for $n=k+1$ is $(k+2)^2+cdots+(2k)^2+(2k+1)^2+(2k+2)^2$.
    – Yadati Kiran
    Nov 24 at 12:38












    Can't i just do $(k+2)^2+...+(2k)^2+(2k+2)^2$
    – Tfue
    Nov 24 at 12:41




    Can't i just do $(k+2)^2+...+(2k)^2+(2k+2)^2$
    – Tfue
    Nov 24 at 12:41












    No. Just put some values for $k$ to get clarity
    – Yadati Kiran
    Nov 24 at 12:42




    No. Just put some values for $k$ to get clarity
    – Yadati Kiran
    Nov 24 at 12:42











    0














    Observe that:



    $$(k+2)^2+cdots+(2k+2)^2=$$$$left[(k+1)^2+(k+2)^2+cdots+(2k)^2right]+(2k+1)^2+(2k+2)^2-(k+1)^2=$$$$f(k)+(2k+1)^2+(2k+2)^2-(k+1)^2$$



    where $f(k)$ denotes the RHS of 2).



    If the equality indeed holds then the outcome must be $f(k+1)$ and that can be checked.






    share|cite|improve this answer




























      0














      Observe that:



      $$(k+2)^2+cdots+(2k+2)^2=$$$$left[(k+1)^2+(k+2)^2+cdots+(2k)^2right]+(2k+1)^2+(2k+2)^2-(k+1)^2=$$$$f(k)+(2k+1)^2+(2k+2)^2-(k+1)^2$$



      where $f(k)$ denotes the RHS of 2).



      If the equality indeed holds then the outcome must be $f(k+1)$ and that can be checked.






      share|cite|improve this answer


























        0












        0








        0






        Observe that:



        $$(k+2)^2+cdots+(2k+2)^2=$$$$left[(k+1)^2+(k+2)^2+cdots+(2k)^2right]+(2k+1)^2+(2k+2)^2-(k+1)^2=$$$$f(k)+(2k+1)^2+(2k+2)^2-(k+1)^2$$



        where $f(k)$ denotes the RHS of 2).



        If the equality indeed holds then the outcome must be $f(k+1)$ and that can be checked.






        share|cite|improve this answer














        Observe that:



        $$(k+2)^2+cdots+(2k+2)^2=$$$$left[(k+1)^2+(k+2)^2+cdots+(2k)^2right]+(2k+1)^2+(2k+2)^2-(k+1)^2=$$$$f(k)+(2k+1)^2+(2k+2)^2-(k+1)^2$$



        where $f(k)$ denotes the RHS of 2).



        If the equality indeed holds then the outcome must be $f(k+1)$ and that can be checked.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 24 at 12:24

























        answered Nov 24 at 11:51









        drhab

        96.5k544127




        96.5k544127























            0














            $(k+2)^2+dots+(2k)^2+(2k+1)^2+(2k+2)^2=frac{k(2k+1)(7k+1)}6-(k+1)^2+(2k+1)^2+(2k+2)^2=frac{k(14k^2+9k+1)+6(7k^2+10k+4)}6=frac{(k+1)(2k+3)(7k+8)}6$.






            share|cite|improve this answer





















            • Hey thanks can you please explain how did you get $(2k+1)^2$ thank you.
              – Tfue
              Nov 25 at 7:03










            • You need all of the $(k+2)^2$ through $(2k+2)^2$.
              – Chris Custer
              Nov 25 at 7:49
















            0














            $(k+2)^2+dots+(2k)^2+(2k+1)^2+(2k+2)^2=frac{k(2k+1)(7k+1)}6-(k+1)^2+(2k+1)^2+(2k+2)^2=frac{k(14k^2+9k+1)+6(7k^2+10k+4)}6=frac{(k+1)(2k+3)(7k+8)}6$.






            share|cite|improve this answer





















            • Hey thanks can you please explain how did you get $(2k+1)^2$ thank you.
              – Tfue
              Nov 25 at 7:03










            • You need all of the $(k+2)^2$ through $(2k+2)^2$.
              – Chris Custer
              Nov 25 at 7:49














            0












            0








            0






            $(k+2)^2+dots+(2k)^2+(2k+1)^2+(2k+2)^2=frac{k(2k+1)(7k+1)}6-(k+1)^2+(2k+1)^2+(2k+2)^2=frac{k(14k^2+9k+1)+6(7k^2+10k+4)}6=frac{(k+1)(2k+3)(7k+8)}6$.






            share|cite|improve this answer












            $(k+2)^2+dots+(2k)^2+(2k+1)^2+(2k+2)^2=frac{k(2k+1)(7k+1)}6-(k+1)^2+(2k+1)^2+(2k+2)^2=frac{k(14k^2+9k+1)+6(7k^2+10k+4)}6=frac{(k+1)(2k+3)(7k+8)}6$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 24 at 12:49









            Chris Custer

            10.5k3724




            10.5k3724












            • Hey thanks can you please explain how did you get $(2k+1)^2$ thank you.
              – Tfue
              Nov 25 at 7:03










            • You need all of the $(k+2)^2$ through $(2k+2)^2$.
              – Chris Custer
              Nov 25 at 7:49


















            • Hey thanks can you please explain how did you get $(2k+1)^2$ thank you.
              – Tfue
              Nov 25 at 7:03










            • You need all of the $(k+2)^2$ through $(2k+2)^2$.
              – Chris Custer
              Nov 25 at 7:49
















            Hey thanks can you please explain how did you get $(2k+1)^2$ thank you.
            – Tfue
            Nov 25 at 7:03




            Hey thanks can you please explain how did you get $(2k+1)^2$ thank you.
            – Tfue
            Nov 25 at 7:03












            You need all of the $(k+2)^2$ through $(2k+2)^2$.
            – Chris Custer
            Nov 25 at 7:49




            You need all of the $(k+2)^2$ through $(2k+2)^2$.
            – Chris Custer
            Nov 25 at 7:49


















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