
Summation Induction help
Hello, I've read everything I can on induction for the most part it makes sense, but I'm having trouble with this summation problem:
SUM from i = 1 to n: (1)^(i+1)(i^2) = [ (1)^(n+1) n(n+1) ] / 2
(I dont know latex...yet)
Basis step:
1 = 1 check
Inductive step:
plugging in n+1 and trying to solve I can't seem to get anywhere...
Please point me in the right direction
Thankyou for any help!!
jeremywho

Well, I'll assume you know how induction is done and how it works. It's just a matter (in this case) of calculation. In fact, suppose the equality holds for some $\displaystyle n \in \mathbb N$, that is:
$\displaystyle \sum_{i=1}^n (1)^{i+1}i^2=(1)^{n+1}\frac{n(n+1)}{2}$
You wish to prove it also holds for $\displaystyle n+1$, hence you want to prove that:
$\displaystyle \sum_{i=1}^{n+1} (1)^{i+1}i^2=(1)^{n+2}\frac{(n+1)(n+2)}{2}$
Simply observe that:
$\displaystyle \sum_{i=1}^{n+1} (1)^{i+1}i^2=(1)^{n+2}(n+1)^2+\sum_{i=1}^{n} (1)^{i+1}i^2$
and recall that you are assuming the equality is true for $\displaystyle n$ (which means you know the corresponding expression for that sum on the right side).
Hope it helped :)

Thankyou so much for the reply!!
I understand we want to prove this:
$\displaystyle
\sum_{i=1}^{n+1} (1)^{i+1}i^2=(1)^{n+2}\frac{(n+1)(n+2)}{2}
$
I don't understand how we get:
$\displaystyle
(1)^{n+2}(n+1)^2
$
in
$\displaystyle
\sum_{i=1}^{n+1} (1)^{i+1}i^2=(1)^{n+2}(n+1)^2+\sum_{i=1}^{n} (1)^{i+1}i^2
$
Maybe its just simple algebra I'm missing?
Thankyou Again!!

You just pick the last term of
$\displaystyle \sum_{i=1}^{n+1} (1)^{i+1}i^2$
which is $\displaystyle (1)^{n+2}(n+1)^2$ and write it independently. A straightforward, numerical example:
$\displaystyle \sum_{i=1}^{5}i = 1+2+3+4+5 = 5+(4+3+2+1) = 5+ (\sum_{i=1}^4i)$
The sums don't have the same upper bound! One ends in $\displaystyle n+1$ and the other in $\displaystyle n$. :p

Got it!
I finally got it! I just wanted to thankyou again for your help!