Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
3  
The formulas are wrong, $2^n$ and $2^{k+1}$ should read $2n$ and $2(k+1)$. –  Did Nov 6 '11 at 19:39

2 Answers 2

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.

share|improve this answer
    
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
4  
@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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.