Prove by induction that $\sum_{k=0}^{n}(-1)^{n+k} k^{2} = \frac{n(n+1)}{2}$ Prove by mathematical induction that 
$\forall n \in \mathbb{N}:~~~~ \sum_{k=0}^{n}(-1)^{n+k} k^{2} = \frac{n(n+1)}{2}$
Step 1: Show true for $n = 1$:
LHS: $(-1)^{(1+0)} \cdot 0^{2} + (-1)^{(1+1)} \cdot 1^{2} = 1$
RHS: $\frac{0(0+1)}{2} + \frac{1(1+1)}{2} = 1$
Step 2: Show "if true for $n = p$, then true for $n = p + 1$":
So we start by the LHS of  $n = p + 1$ and try to get to the RHS of it, by using the equality for $n = p$. 
The first step involves breaking out the highest term in the sum.
$\sum_{k=0}^{p+1}(-1)^{(p+1+k)} k^{2} = (-1)^{(2p+2)} (p+1)^{2} + \sum_{k=0}^{p}(-1)^{(p+k)} k^{2} $
Replacing the last term with the equality for $n = p$.
$(-1)^{2p+2} (p+1)^{2} + \sum_{k=0}^{n}(-1)^{p+k} k^{2} = (-1)^{2p+2} (p+1)^{2} + \frac{p(p+1)}{2}$
Since 2p+2 is an even number, the -1 becomes a 1:
$(-1)^{2p+2} (p+1)^{2} + \frac{p(p+1)}{2} = (p+1)^{2} + \frac{p(p+1)}{2}$
All that is required now is to show that this equals to $\frac{(p+1)(p+2)}{2}$, but there seems to be a $(p+1)$ too many in the first term for it to work.
Question 1: Where have I gone wrong in the above attempt?
Question 2: Is it more productive to attempt to prove 
$\forall n \in \mathbb{N}:~~~~ \sum_{k=0}^{n}(-1)^{n+k} k^{2} = \sum_{k=1}^{n} k$
instead?

In this step you have made an error:
$$\sum_{k=0}^{p+1}(-1)^{(p+1+k)} k^{2} = (-1)^{(2p+2)} (p+1)^{2} + \sum_{k=0}^{p}(-1)^{(p+k)} k^{2}$$
This should be 
$$\sum_{k=0}^{p+1}(-1)^{(p+1+k)} k^{2} = (-1)^{(2p+2)} (p+1)^{2} + \sum_{k=0}^{p}(-1)^{(p+1+k)} k^{2}.$$Notice the extra $+1$ in the exponent of $(-1)$ in the last sum. We can remove it by factoring out a $-1$ to get
$$\sum_{k=0}^{p+1}(-1)^{(p+1+k)} k^{2} =  (p+1)^{2} - \sum_{k=0}^{p}(-1)^{(p+k)} k^{2}.$$
By the assumption this is
$$\sum_{k=0}^{p+1}(-1)^{(p+1+k)} k^{2} =  (p+1)^{2} -\frac{p(p+1)}{2}.$$
Now we can easily work it out:
$$(p+1)^2 - \frac{p(p+1)}{2} = (p+1)(-\frac{p}{2} + p +1)= (p+1)\frac{p+2}{2}$$
as desired.
