Is my proof by induction of even number arithmetic progression correct?












1












$begingroup$


Prove that for all integers $ngeq1$, $P(n) = 2+4+6+...+2n=n^2+n$



For $P(1)$, we have that: $2cdot1=1^2+1$



Let $k$ be an integer with $kgeq1$. Suppose $P(k)$ is true. Then: $2+4+6+...+2k=k^2+k$



I must show that $P(k+1)$ is also true:



$2+4+6+...+2(k+1)=(k+1)^2+(k+1)$ $(1)$



If $k$ is an integer then $l=k+1$ is also an integer. For $l=k+1$ $(1)$ becomes: $l^2+l$ which is of the form $k^2+k$ that is true.



$Q.E.D$



A small change at the inductive step:



I must show that $P(k+1)$ is also true:



$(2+4+6+...+2k)+2(k+1)=k^2+k+2k+2=(k^2+2k+1)+(k+1)=(k+1)^2+(k+1)$



$Q.E.D$



I know that the second inductive step is definitely correct, but I'm confused as to whether the first one is correct.






discrete-mathematics induction







share|cite|improve this question









$endgroup$












  • $begingroup$
    I think it's right.
    $endgroup$
    – Michael Rozenberg
    Jan 5 at 13:07






  • 3




    $begingroup$
    First one is wrong.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:07






  • 3




    $begingroup$
    I don't understand the first argument at all. It looks like you just assumed what you were trying to prove. The second one looks solid.
    $endgroup$
    – lulu
    Jan 5 at 13:09






  • 1




    $begingroup$
    You can't make arguments using equation $(1) $,as you haven't proved it.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:10










  • $begingroup$
    The first attempt isn't valid.
    $endgroup$
    – Peter
    Jan 5 at 13:14
















1












$begingroup$


Prove that for all integers $ngeq1$, $P(n) = 2+4+6+...+2n=n^2+n$



For $P(1)$, we have that: $2cdot1=1^2+1$



Let $k$ be an integer with $kgeq1$. Suppose $P(k)$ is true. Then: $2+4+6+...+2k=k^2+k$



I must show that $P(k+1)$ is also true:



$2+4+6+...+2(k+1)=(k+1)^2+(k+1)$ $(1)$



If $k$ is an integer then $l=k+1$ is also an integer. For $l=k+1$ $(1)$ becomes: $l^2+l$ which is of the form $k^2+k$ that is true.



$Q.E.D$



A small change at the inductive step:



I must show that $P(k+1)$ is also true:



$(2+4+6+...+2k)+2(k+1)=k^2+k+2k+2=(k^2+2k+1)+(k+1)=(k+1)^2+(k+1)$



$Q.E.D$



I know that the second inductive step is definitely correct, but I'm confused as to whether the first one is correct.






discrete-mathematics induction







share|cite|improve this question









$endgroup$












  • $begingroup$
    I think it's right.
    $endgroup$
    – Michael Rozenberg
    Jan 5 at 13:07






  • 3




    $begingroup$
    First one is wrong.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:07






  • 3




    $begingroup$
    I don't understand the first argument at all. It looks like you just assumed what you were trying to prove. The second one looks solid.
    $endgroup$
    – lulu
    Jan 5 at 13:09






  • 1




    $begingroup$
    You can't make arguments using equation $(1) $,as you haven't proved it.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:10










  • $begingroup$
    The first attempt isn't valid.
    $endgroup$
    – Peter
    Jan 5 at 13:14














1












1








1





$begingroup$


Prove that for all integers $ngeq1$, $P(n) = 2+4+6+...+2n=n^2+n$



For $P(1)$, we have that: $2cdot1=1^2+1$



Let $k$ be an integer with $kgeq1$. Suppose $P(k)$ is true. Then: $2+4+6+...+2k=k^2+k$



I must show that $P(k+1)$ is also true:



$2+4+6+...+2(k+1)=(k+1)^2+(k+1)$ $(1)$



If $k$ is an integer then $l=k+1$ is also an integer. For $l=k+1$ $(1)$ becomes: $l^2+l$ which is of the form $k^2+k$ that is true.



$Q.E.D$



A small change at the inductive step:



I must show that $P(k+1)$ is also true:



$(2+4+6+...+2k)+2(k+1)=k^2+k+2k+2=(k^2+2k+1)+(k+1)=(k+1)^2+(k+1)$



$Q.E.D$



I know that the second inductive step is definitely correct, but I'm confused as to whether the first one is correct.






discrete-mathematics induction







share|cite|improve this question









$endgroup$




Prove that for all integers $ngeq1$, $P(n) = 2+4+6+...+2n=n^2+n$



For $P(1)$, we have that: $2cdot1=1^2+1$



Let $k$ be an integer with $kgeq1$. Suppose $P(k)$ is true. Then: $2+4+6+...+2k=k^2+k$



I must show that $P(k+1)$ is also true:



$2+4+6+...+2(k+1)=(k+1)^2+(k+1)$ $(1)$



If $k$ is an integer then $l=k+1$ is also an integer. For $l=k+1$ $(1)$ becomes: $l^2+l$ which is of the form $k^2+k$ that is true.



$Q.E.D$



A small change at the inductive step:



I must show that $P(k+1)$ is also true:



$(2+4+6+...+2k)+2(k+1)=k^2+k+2k+2=(k^2+2k+1)+(k+1)=(k+1)^2+(k+1)$



$Q.E.D$



I know that the second inductive step is definitely correct, but I'm confused as to whether the first one is correct.






discrete-mathematics induction




discrete-mathematics induction






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 5 at 13:04









NickNick

183




183












  • $begingroup$
    I think it's right.
    $endgroup$
    – Michael Rozenberg
    Jan 5 at 13:07






  • 3




    $begingroup$
    First one is wrong.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:07






  • 3




    $begingroup$
    I don't understand the first argument at all. It looks like you just assumed what you were trying to prove. The second one looks solid.
    $endgroup$
    – lulu
    Jan 5 at 13:09






  • 1




    $begingroup$
    You can't make arguments using equation $(1) $,as you haven't proved it.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:10










  • $begingroup$
    The first attempt isn't valid.
    $endgroup$
    – Peter
    Jan 5 at 13:14


















  • $begingroup$
    I think it's right.
    $endgroup$
    – Michael Rozenberg
    Jan 5 at 13:07






  • 3




    $begingroup$
    First one is wrong.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:07






  • 3




    $begingroup$
    I don't understand the first argument at all. It looks like you just assumed what you were trying to prove. The second one looks solid.
    $endgroup$
    – lulu
    Jan 5 at 13:09






  • 1




    $begingroup$
    You can't make arguments using equation $(1) $,as you haven't proved it.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:10










  • $begingroup$
    The first attempt isn't valid.
    $endgroup$
    – Peter
    Jan 5 at 13:14
















$begingroup$
I think it's right.
$endgroup$
– Michael Rozenberg
Jan 5 at 13:07




$begingroup$
I think it's right.
$endgroup$
– Michael Rozenberg
Jan 5 at 13:07




3




3




$begingroup$
First one is wrong.
$endgroup$
– Thomas Shelby
Jan 5 at 13:07




$begingroup$
First one is wrong.
$endgroup$
– Thomas Shelby
Jan 5 at 13:07




3




3




$begingroup$
I don't understand the first argument at all. It looks like you just assumed what you were trying to prove. The second one looks solid.
$endgroup$
– lulu
Jan 5 at 13:09




$begingroup$
I don't understand the first argument at all. It looks like you just assumed what you were trying to prove. The second one looks solid.
$endgroup$
– lulu
Jan 5 at 13:09




1




1




$begingroup$
You can't make arguments using equation $(1) $,as you haven't proved it.
$endgroup$
– Thomas Shelby
Jan 5 at 13:10




$begingroup$
You can't make arguments using equation $(1) $,as you haven't proved it.
$endgroup$
– Thomas Shelby
Jan 5 at 13:10












$begingroup$
The first attempt isn't valid.
$endgroup$
– Peter
Jan 5 at 13:14




$begingroup$
The first attempt isn't valid.
$endgroup$
– Peter
Jan 5 at 13:14










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