How was the closed form of this alternating sum of squares calculated? I am reading through this answer at socratic.org.
The question is to find the closed form of the sum
$$1^{2}-2^{2}+3^{2}-4^{2}+5^{2}-6^{2}+\ldots.$$
I understand that, if the terms were added, the sum would be
$$
\sum_{n=1}^{N} n^{2}=1^{2}+2^{2}+\ldots+N^{2}.
$$
The person goes on to say, if the series were not alternating, the sum would be
$$
S=\frac{N(N+1)}{2}
$$
But is that correct? I thought the sum of the first $N$ squares would be
$$
\frac{N(N+1)(2N+1)}{6}.
$$
Lastly, I understand moving the $-1$ constant out of the summation as such
$$
=-\sum_{n=1}^{N}(-1)^{n} n^{2}
$$
But I am completely missing how the final closed form was calculated
$$
S_{N}=-\frac{(-1)^{N} N(N+1)}{2}
$$
 A: It is indeed not true that $\sum_{n=1}^Nn^2=\frac{N(N+1)}{2}$ for each positive integer $N$. This is easily verified by plugging in any value $N\geq2$. Of course it is true that $\sum_{n=1}^Nn=\frac{N(N+1)}{2}$. And you are indeed correct when you say that
$$\sum_{n=1}^Nn^2=\frac{N(N+1)(2N+1)}{6}.$$
Do note that if $N$ is even, say $N=2M$, then
\begin{eqnarray*}
\sum_{n=1}^N(-1)^{n+1}n^2
&=&\sum_{n=1}^M\big((-1)^{2n}(2n-1)^2+(-1)^{2n+1}(2n)^2\big).
\end{eqnarray*}
Of course $(-1)^{2n}=1$ for all $n$, and similarly $(-1)^{2n+1}=-1$. It follows that for all $n$ we have
\begin{eqnarray*}
(-1)^{2n}(2n-1)^2+(-1)^{2n+1}(2n)^2
&=&(2n-1)^2-(2n)^2\\
&=&(4n^2-4n+1)-4n^2\\
&=&1-4n.
\end{eqnarray*}
This shows that
$$\sum_{n=1}^N(-1)^{n+1}n^2=\sum_{n=1}^M(1-4n).$$
From here we can use the fact that $\sum_{n=1}^Mn=\frac{M(M+1)}{2}$ to find that
\begin{eqnarray*}
\sum_{n=1}^M(1-4n)&=&M-4\sum_{n=1}^Mn\\
&=&M-4\cdot\frac{M(M+1)}{2}\\
&=&-2M^2-M\\
&=&-\frac{N(N+1)}{2}.
\end{eqnarray*}
That proves the case where $N$ is even. If $N$ is odd, say $N=2M+1$, then
\begin{eqnarray*}
\sum_{n=1}^N(-1)^{n+1}n^2
&=&\left(\sum_{n=1}^{2M}(-1)^{n+1}n^2\right)+N^2\\
&=&-\frac{(N-1)N}{2}+N^2\\
&=&\frac{N(N+1)}{2}.
\end{eqnarray*}
This shows that although the reasoning is incorrect, the conclusion does indeed hold; that
$$\sum_{n=1}^N(-1)^{n+1}n^2=-(-1)^N\frac{N(N+1)}{2}.$$

Alternatively, you could note that
$$1^2-2^2+3^2-4^2+5^2-\ldots=(1^2+2^2+3^2+4^2+5^2+\ldots)-2(2^2+4^2+8^2+10^2+\ldots).$$
So from your observation that
$$\sum_{n=1}^Nn^2=\frac{N(N+1)(2N+1)}{6},$$
you could conclude that if $N$ is even, say $N=2M$, then
\begin{eqnarray*}
\sum_{n=1}^N(-1)^{n+1}n^2
&=&\left(\sum_{n=1}^{2M}n^2\right)-2\left(\sum_{n=1}^M(2n)^2\right)\\
&=&\frac{2M(2M+1)(4M+1)}{6}-8\frac{M(M+1)(2M+1)}{6}\\
&=&\frac{-12M^2-6M}{6}\\
&=&-\frac{N(N+1)}{2}.
\end{eqnarray*}
A: Without analyzing odd and even cases, we can also construct the closed-form formula as follows:
Using the formula $S_n-S_{n-1}=a_n$,  where $a_n=(-1)^{n+1}n^2$ and define $S_n=(-1)^n\left(an^2+bn\right)$,  then we have
$$\begin{align}S_n-S_{n-1}=(-1)^n(an^2+bn)+(-1)^{n}\left(a(n-1)^2+b(n-1)\right)=2a(-1)^nn^2-2n(a-b)+(a-b)=-(-1)^nn^2\end{align}$$
This implies,
$$a=b=-\frac 12$$
This means,
$$\begin{align}\sum^{n}_{k=1} (-1)^{k+1}k^2
&=(-1)^n\left(-\frac 12n^2-\frac 12n\right)\\
&=\frac 12(-1)^{n+1}n(n+1).\end{align}$$
A: (This answer is an expansion of my comment. It shows an alternative method for computing the sum.)
If $n$ is even, then an easy way to compute $1^2-2^2+3^2-4^2+\dots+(n-1)^2+n^2$ is to consider the difference between each pair of terms:
\begin{align}
&1^2-2^2+3^2-4^2+\dots+(n-1)^2+n^2 \\[4pt]
=&(1-2)(1+2)+(3-4)(3+4)+\dots +((n-1)-n)((n-1)+n) \\[4pt]
=&-1(1+2)-1(3+4)-\dots-1((n-1)+n) \\[4pt]
=&-1(1+2+3+4+\dots+(n-1)+n) \\[4pt]
=&-\frac{n(n+1)}{2}
\end{align}
If $n$ is odd, then $n-1$ is even, and so we can use the above formula:
$$
1^2-2^2+3^2-4^2+\dots+(n-2)^2+(n-1)^2=-\frac{(n-1)n}{2} \, .
$$
Then, adding $n^2$ to both sides, we find that
$$
1^2-2^2+3^2-4^2+\dots+(n-1)^2+n^2=\frac{n(n+1)}{2} \, .
$$
Therefore, regardless of the parity of $n$,
$$
1^2-2^2+3^2-4^2+\dots+(n-1)^2+n^2=(-1)^{n+1}\cdot\frac{n(n+1)}{2} \, .
$$
A: A quick intuition :
If $N$ is odd
$\displaystyle S = -\sum_{n=1}^{N} (-1)^{n} n^2  = 1 + (-2^2 + 3^2) + ... + (-1)^{N+1}(-(N-1)^2  + N^2)$
Hence $\displaystyle S = 1  +  5 + 9 + ... + (2N-1)  $
There are $\frac {N+1}{2} $ terms in the above expression ( which is an A.P. in with common difference of four).
Thus $\displaystyle S_{N} = \frac {N+1}{2} \cdot \frac {2N-1+1}{2} = \frac {N(N+1)}{2}  $
Similarly when $N$ is even then the sum till  $N-1$ is $\displaystyle S_{N-1} =  \frac {(N-1)N}{2}  $
Add $-N^2$ to it to get $\displaystyle S_{N} $ when N is even :  $\displaystyle S_{N} =  \frac {(N-1)N}{2} - N^2 = -\frac {(N+1)N}{2}  $
Hope this helps, have edited the errors.
