5

I need to prove by induction this thing: $2+4+6+........+2n = n(n+1)$

so, this thing is composed by sum of pair numbers, so its what I do, but I'm stucked.

$2+4+6+\cdots+2n = n(n+1)$

$(2+4+6+\cdots+2n)+(2n+2) = n(n+1) + (2n+2) $

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

$n^2 + 3n + 2$

$n(n+2+1)+2$

I don't know how to move forward from this.

Thanks.

Kevin
  • 567
PlayMa256
  • 711
  • 1
    You were almost there. When working with induction, I suggest you write what both sides should look like for $n+1$. You would've noticed the factorization of your second to last equation completed the proof: $n^2+3n+2=(n+1)(n+2)$ – Kevin Nov 27 '13 at 22:42

3 Answers3

7

$2+4+6+........+2n = n(n+1)$

$2+4+6+........+2n+2(n+1) = n(n+1)+2(n+1)=(n+1)(n+2)$

Adi Dani
  • 16,949
5

Suppose that $$ 2+4+6+\cdots+2n=n(n+1) $$ add $2n+2$ to both sides: $$ 2+4+6+\cdots+2n+(2n+2)=n(n+1)+(2n+2) $$ To finish the induction, you want the right side to be $(n+1)(n+2)$. Is it? $$ n(n+1)+(2n+2)=n^2+n+2n+2=n^2+3n+2=(n+1)(n+2) $$

robjohn
  • 345,667
3

The statement is: $2+4+...+2n=n(n+1)$ for all $n$

Now, $n^2+3n+2=(n+1)(n+2)=(n+1)((n+1)+1)$ and that is what you want to have as a the result for $n+1$.

Stefan Hamcke
  • 27,733