Proof alternating sum of squares is alternating sign of sum I'm trying to prove by induction that $1-4+9-...\pm n^2 = \pm(1+2+...+n)$.  The base-case is obvious, and the formula that I write this as is 
$$\sum_{i=1}^{n}(-1)^{i+1}i^{2} = (-1)^{n+1}\sum_{i=1}^{n}i$$
If we assume this holds up to $n$ then I try to prove the equation true for $n+1$ but get stuck.  I have 
$$\sum_{i=1}^{n+1}(-1)^{i+1}i^{2} = (-1)^{n+1}\sum_{i=1}^{n}i + (-1)^{n+2}(n+1)^{2}$$
$$ = (-1)^{n+2}\Big( (n+1)^{2}-\sum_{i=1}^{n}i\Big)$$
At this point I can't think of anyth
ing that sounds like a good idea.  I suppose I could try to use what I have to get a formula for $n+1$ like moving everything away from this in the inductive hypothesis, $n+1=\pm(1-4+...\pm n^{2})-2-3-...-(n-1)$ but that sounds like I'll just be heading in a circle.
For guidance I can try setting this equal to my goal and reduce it to a known equation, but that doesn't seem to help.
$$(-1)^{n+2}\Big((n+1)^{2}-\sum_{i=1}^{n}i\Big) = (-1)^{n+2}\sum_{i=1}^{n+1}i$$
iff 
$$n^{2}+2n+1 = n+1+2\sum_{i=1}^{n}i$$
iff
$$n^{2} = n+ 2\sum_{i=1}^{n-1}i$$
but this is not obviously true and I can't see where to go from here either.
 A: Let's assume $1^2 - 2^2 + 3^2 - \ldots + (n-1)^2 = 1 + 2 + 3 + \ldots + (n - 1)$, and see what happens when we subtract $n^2$. Note that subtraction must come next, as we're assuming we have a positive sum.
To use this proof, you must be able to use the fact that $1 + 2 + \ldots + (n-1) = \frac{n(n-1)}{2}$.
Subtracting $n^2$, we get $$\frac{n(n - 1)}{2} - n^2 = \frac{n^2 - n}{2} - \frac{2n^2}{2} = \frac{-n^2 - n}{2} = -\frac{n(n+1)}{2} = -(1 + 2 + \ldots + n).$$
Does that help?
A: Recall that $$\sum_{k = 1}^{n} k = \frac{n(n + 1)}{2}$$
For the induction step:  Let the induction hypothesis be 
$$\sum_{k = 1}^{m} (-1)^{k + 1} k^2 = (-1)^{m + 1}\sum_{k = 1}^{m} k$$
for some $m \in \mathbb{N}$.  Let $n = m + 1$.  Then 
\begin{align*}
\sum_{k = 1}^{m + 1} (-1)^{k + 1} k^2 & = \sum_{k = 1}^{m} (-1)^{k + 1} k^2 + (-1)^{m + 2} (m + 1)^2\\
& = (-1)^{m + 1}\sum_{k = 1}^{m} k + (-1)^{m + 2} (m + 1)^2 && \text{induction hypothesis}\\
& = (-1)^{m + 1} \frac{m(m + 1)}{2} + (-1)^{m + 2} (m + 1)^2\\
& = (-1)^{m + 2}(m + 1)\left[-\frac{m}{2} + m + 1\right]\\
& = (-1)^{m + 2}\frac{m + 1}{2}(-m + 2m + 2)\\
& = (-1)^{m + 2}\frac{(m + 1)(m + 2)}{2}\\
& = (-1)^{m + 2}\sum_{k = 1}^{m + 1} k
\end{align*}
A: A short answer to your question:
Note that $-r^2+(r+1)^2-r^2=r+(r+1)$.
Hence, for odd $n$, 
$$\begin{align}
&1^2
\underbrace{-2^2+3^2}_{2+3}
\underbrace{-4^2+5^2}_{4+5}+\cdots+\underbrace{-(n-1)^2+n^2}_{(n-1)+n}
&=1+2+3+4+\cdots+n\\
\end{align}$$
and for even $n$,
$$\begin{align}
&\underbrace{1^2-2^2}_{-1-2}+
\underbrace{3^2-4^2}_{-3-4}+\cdots+\underbrace{(n-1)^2-n^2}_{-(n-1)-n}
&=-\bigg(1+2+3+4+\cdots+n\bigg)
\end{align}$$ 
.  
In summary, 
$$1^2-2^2+3^3-4^4+\cdots+(-1)^{n-1} n^2
=(-1)^{n-1}\bigg(1+2+3+4+\cdots+n\bigg)\;\blacksquare$$
or
$$\sum_{r=1}^n (-1)^{r-1}r^2=(-1)^{n-1}\sum_{r=1}^n r=(-1)^{n-1}\binom {n+1}2$$
