Proving $(n+1)^2+(n+2)^2+…+(2n)^2= frac{n(2n+1)(7n+1)}{6}$ by induction
My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= frac{n(2n+1)(7n+1)}{6}$
My workings
- LHS=$2^2$ =$4$ RHS= $frac{24}{6} =4 $
- $(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= frac{k(2k+1)(7k+1)}{6}$
- 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
edited Nov 24 at 13:39
user21820
38.6k542153
asked Nov 24 at 11:38
Tfue
859
|
show 4 more comments
My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= frac{n(2n+1)(7n+1)}{6}$
My workings
- LHS=$2^2$ =$4$ RHS= $frac{24}{6} =4 $
- $(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= frac{k(2k+1)(7k+1)}{6}$
- 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
edited Nov 24 at 13:39
user21820
38.6k542153
asked Nov 24 at 11:38
Tfue
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
|
show 4 more comments
My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= frac{n(2n+1)(7n+1)}{6}$
My workings
- LHS=$2^2$ =$4$ RHS= $frac{24}{6} =4 $
- $(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= frac{k(2k+1)(7k+1)}{6}$
- 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
edited Nov 24 at 13:39
user21820
38.6k542153
asked Nov 24 at 11:38
Tfue
859
My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= frac{n(2n+1)(7n+1)}{6}$
My workings
- LHS=$2^2$ =$4$ RHS= $frac{24}{6} =4 $
- $(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= frac{k(2k+1)(7k+1)}{6}$
- 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
edited Nov 24 at 13:39
user21820
38.6k542153
asked Nov 24 at 11:38
Tfue
859
edited Nov 24 at 13:39
user21820
38.6k542153
asked Nov 24 at 11:38
Tfue
859
edited Nov 24 at 13:39
user21820
38.6k542153
edited Nov 24 at 13:39
user21820
38.6k542153
edited Nov 24 at 13:39
user21820
38.6k542153
38.6k542153
asked Nov 24 at 11:38
Tfue
859
asked Nov 24 at 11:38
Tfue
859
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
|
show 4 more comments
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
|
show 4 more comments
5 Answers
5
active
oldest
votes
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:
$$(n+1)^2+(n+2)^2+ldots+(n+n)^2=$$
$$=n^2+ldots+n^2+2n+4n+ldots+2nn+1+2^2+ldots+n^2=$$
$$ncdot n^2+2(1+2+ldots+n)n+frac{n(n+1)(2n+1)}6=$$
$$=n^3+n^2(n+1)+frac{n(n+1)(2n+1)}6=ldots$$
answered Nov 24 at 11:58
DonAntonio
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
add a comment |
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
answered Nov 24 at 11:53
Martund
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
add a comment |
Assuming the step for $n=k$ $:$For
$n=k+1$ $(k+2)^2+cdots+(2k+2)^2=dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2$ $=dfrac{k(2k+1)(7k+1)}{6}+3(k+1)^2+(2k+1)^2=dfrac{k(2k+1)(7k+1)}{6}+7k^2+10k+4=dfrac{(k+1)(2k+3)(7k+8)}{6}$
answered Nov 24 at 11:54
Yadati Kiran
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
add a comment |
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.
answered Nov 24 at 11:51
drhab
96.5k544127
add a comment |
$(k+2)^2+dots+(2k)^2+(2k+1)^2+(2k+2)^2=frac{k(2k+1)(7k+1)}6-(k+1)^2+(2k+1)^2+(2k+2)^2=frac{k(14k^2+9k+1)+6(7k^2+10k+4)}6=frac{(k+1)(2k+3)(7k+8)}6$.
answered Nov 24 at 12:49
Chris Custer
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
add a comment |
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5 Answers
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5 Answers
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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:
$$(n+1)^2+(n+2)^2+ldots+(n+n)^2=$$
$$=n^2+ldots+n^2+2n+4n+ldots+2nn+1+2^2+ldots+n^2=$$
$$ncdot n^2+2(1+2+ldots+n)n+frac{n(n+1)(2n+1)}6=$$
$$=n^3+n^2(n+1)+frac{n(n+1)(2n+1)}6=ldots$$
answered Nov 24 at 11:58
DonAntonio
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
add a comment |
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:
$$(n+1)^2+(n+2)^2+ldots+(n+n)^2=$$
$$=n^2+ldots+n^2+2n+4n+ldots+2nn+1+2^2+ldots+n^2=$$
$$ncdot n^2+2(1+2+ldots+n)n+frac{n(n+1)(2n+1)}6=$$
$$=n^3+n^2(n+1)+frac{n(n+1)(2n+1)}6=ldots$$
answered Nov 24 at 11:58
DonAntonio
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
add a comment |
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:
$$(n+1)^2+(n+2)^2+ldots+(n+n)^2=$$
$$=n^2+ldots+n^2+2n+4n+ldots+2nn+1+2^2+ldots+n^2=$$
$$ncdot n^2+2(1+2+ldots+n)n+frac{n(n+1)(2n+1)}6=$$
$$=n^3+n^2(n+1)+frac{n(n+1)(2n+1)}6=ldots$$
answered Nov 24 at 11:58
DonAntonio
176k1491225
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:
$$(n+1)^2+(n+2)^2+ldots+(n+n)^2=$$
$$=n^2+ldots+n^2+2n+4n+ldots+2nn+1+2^2+ldots+n^2=$$
$$ncdot n^2+2(1+2+ldots+n)n+frac{n(n+1)(2n+1)}6=$$
$$=n^3+n^2(n+1)+frac{n(n+1)(2n+1)}6=ldots$$
answered Nov 24 at 11:58
DonAntonio
176k1491225
edited Nov 24 at 12:08
answered Nov 24 at 11:58
DonAntonio
176k1491225
answered Nov 24 at 11:58
DonAntonio
176k1491225
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
add a comment |
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
add a comment |
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
answered Nov 24 at 11:53
Martund
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
add a comment |
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
answered Nov 24 at 11:53
Martund
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
add a comment |
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
answered Nov 24 at 11:53
Martund
1,344212
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
answered Nov 24 at 11:53
Martund
1,344212
answered Nov 24 at 11:53
Martund
1,344212
answered Nov 24 at 11:53
Martund
1,344212
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
add a comment |
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
add a comment |
Assuming the step for $n=k$ $:$For
$n=k+1$ $(k+2)^2+cdots+(2k+2)^2=dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2$ $=dfrac{k(2k+1)(7k+1)}{6}+3(k+1)^2+(2k+1)^2=dfrac{k(2k+1)(7k+1)}{6}+7k^2+10k+4=dfrac{(k+1)(2k+3)(7k+8)}{6}$
answered Nov 24 at 11:54
Yadati Kiran
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
add a comment |
Assuming the step for $n=k$ $:$For
$n=k+1$ $(k+2)^2+cdots+(2k+2)^2=dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2$ $=dfrac{k(2k+1)(7k+1)}{6}+3(k+1)^2+(2k+1)^2=dfrac{k(2k+1)(7k+1)}{6}+7k^2+10k+4=dfrac{(k+1)(2k+3)(7k+8)}{6}$
answered Nov 24 at 11:54
Yadati Kiran
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
add a comment |
Assuming the step for $n=k$ $:$For
$n=k+1$ $(k+2)^2+cdots+(2k+2)^2=dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2$ $=dfrac{k(2k+1)(7k+1)}{6}+3(k+1)^2+(2k+1)^2=dfrac{k(2k+1)(7k+1)}{6}+7k^2+10k+4=dfrac{(k+1)(2k+3)(7k+8)}{6}$
answered Nov 24 at 11:54
Yadati Kiran
1,688519
Assuming the step for $n=k$ $:$For
$n=k+1$ $(k+2)^2+cdots+(2k+2)^2=dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2$ $=dfrac{k(2k+1)(7k+1)}{6}+3(k+1)^2+(2k+1)^2=dfrac{k(2k+1)(7k+1)}{6}+7k^2+10k+4=dfrac{(k+1)(2k+3)(7k+8)}{6}$
answered Nov 24 at 11:54
Yadati Kiran
1,688519
answered Nov 24 at 11:54
Yadati Kiran
1,688519
answered Nov 24 at 11:54
Yadati Kiran
1,688519
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
add a comment |
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
add a comment |
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.
answered Nov 24 at 11:51
drhab
96.5k544127
add a comment |
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.
answered Nov 24 at 11:51
drhab
96.5k544127
add a comment |
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.
answered Nov 24 at 11:51
drhab
96.5k544127
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.
answered Nov 24 at 11:51
drhab
96.5k544127
edited Nov 24 at 12:24
answered Nov 24 at 11:51
drhab
96.5k544127
answered Nov 24 at 11:51
drhab
96.5k544127
answered Nov 24 at 11:51
drhab
96.5k544127
96.5k544127
add a comment |
add a comment |
$(k+2)^2+dots+(2k)^2+(2k+1)^2+(2k+2)^2=frac{k(2k+1)(7k+1)}6-(k+1)^2+(2k+1)^2+(2k+2)^2=frac{k(14k^2+9k+1)+6(7k^2+10k+4)}6=frac{(k+1)(2k+3)(7k+8)}6$.
answered Nov 24 at 12:49
Chris Custer
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
add a comment |
$(k+2)^2+dots+(2k)^2+(2k+1)^2+(2k+2)^2=frac{k(2k+1)(7k+1)}6-(k+1)^2+(2k+1)^2+(2k+2)^2=frac{k(14k^2+9k+1)+6(7k^2+10k+4)}6=frac{(k+1)(2k+3)(7k+8)}6$.
answered Nov 24 at 12:49
Chris Custer
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
add a comment |
$(k+2)^2+dots+(2k)^2+(2k+1)^2+(2k+2)^2=frac{k(2k+1)(7k+1)}6-(k+1)^2+(2k+1)^2+(2k+2)^2=frac{k(14k^2+9k+1)+6(7k^2+10k+4)}6=frac{(k+1)(2k+3)(7k+8)}6$.
answered Nov 24 at 12:49
Chris Custer
10.5k3724
$(k+2)^2+dots+(2k)^2+(2k+1)^2+(2k+2)^2=frac{k(2k+1)(7k+1)}6-(k+1)^2+(2k+1)^2+(2k+2)^2=frac{k(14k^2+9k+1)+6(7k^2+10k+4)}6=frac{(k+1)(2k+3)(7k+8)}6$.
answered Nov 24 at 12:49
Chris Custer
10.5k3724
answered Nov 24 at 12:49
Chris Custer
10.5k3724
answered Nov 24 at 12:49
Chris Custer
10.5k3724
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
add a comment |
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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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