# $2+4+6+\cdots+2n=n^2+n$ by mathematical induction

I am trying to prove $2+4+6+\cdots +2n=n^2+n$ by mathematical induction. I followed all the steps and the $P_{k+1}$ was $2+4+6+\cdots+2(k+1)=(k+1)^2+k+1$ Starting from the left hand side of the equation I have solved till $k^2+k+2(k+1)$. Now I am stuck here. I don't know how to do it further. Please guide me thanks.

-
The formulas are wrong, $2^n$ and $2^{k+1}$ should read $2n$ and $2(k+1)$. – Did Nov 6 '11 at 19:39

It looks as if you need $2+4+6+\cdots+2n$ rather than $2+4+6+\cdots+2^n$.

Look at $$\underbrace{2+4+6+\cdots+2k}+2(k+1).$$ The induction hypothesis tells you what to do with the part over the underbrace. Then massage it a bit with some simple algebra.

-
No, its the 2nd one which I want... – Fahad Uddin Nov 6 '11 at 19:59
$2+4+6+8+10+12+14+16= 2+4+6+\cdots+2^4$. Is this equal to $4^2+4$? If you go from $4$ to $5$, you get $2+4+6+8+10+\cdots+32$, and you've got 8 more terms, not just one more term. – Michael Hardy Nov 6 '11 at 20:10
@Akito If you're trying to prove that $2+4+\ldots+2^n=n^2+n$, it's not surprising that you're having trouble: it's wrong, you won't be able to prove it. Notice that $6$ is $2\times3$, not $2^3$. – Gilles Nov 6 '11 at 20:14
@Gilles: Thanks for pointing it. Its a wrong proposition – Fahad Uddin Nov 7 '11 at 15:17

$k^2+k+2(k+1)=k^2+3k+2=k^2+2k+1+(k+1)= (k+1)^2+(k+1)$

-