Sum of terms $1^2-2^2+3^2-...$ Prove that: $$1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}{2}$$
I proved it by induction but is there any other way to solve it?
If it was not a proof but rather a question like find the term,how to solve it?
I realized that alternate terms were under same sign but can't understand whether to take $\frac{n}{2}$ odd and even terms[if n is even] or $\frac{n+1}{2}$ odd terms and $\frac{n-1}{2}$ even terms[if n is odd]. 
I thought of this $$1^2-2^2+3^2-4^2+5^2=1^2+{(1+2)}^2+{(1+4)}^2-2^2-4^2$$
$$=1^2+1^2+1^2+2(0+2+4)+2^2+4^2-2^2-4^2$$
But then how to generalize??
 A: Notice $$n^2 = \frac{(n-1)n + n(n+1)}{2}
\quad\implies\quad (-1)^{n-1} n^2 = (-1)^{n-1}\frac{(n-1)n}{2} - (-1)^{n}\frac{n(n+1)}{2}$$
The sum is a telescoping sum and
$$\require{cancel}\sum_{k=1}^n (-1)^{k-1} k^2 = \color{red}{\cancelto{0}{\color{grey}{(-1)^{1-1}\frac{(1-1)1}{2}}}}- (-1)^n \frac{n(n+1)}{2}
= (-1)^{n-1}\frac{n(n+1)}{2}
$$
A: You could use formal manipulations with a truncated power series:
$$\frac{1-x^{n+1}}{1-x} = 1 + x + x^2 + \cdots + x^n \\
\frac{d}{dx} \left( \frac{1-x^{n+1}}{1-x} \right) = 1 + 2x + \cdots + n x^{n-1} \\
x \frac{d}{dx} \left( \frac{1-x^{n+1}}{1-x} \right) = x + 2x^2 + \cdots + n x^n \\
\frac{d}{dx} \left( x \frac{d}{dx} \left( \frac{1-x^{n+1}}{1-x} \right) \right) = 1 + 4x + \cdots + n^2 x^{n-1}
$$
Now evaluate the derivatives on the left hand side, and then substitute $x := -1$.
A: hint: A classic trick is for example $1^2 - 2^2 + 3^2 - 4^2 = (1^2 +2^2+3^2+4^2) - 2(2^2+4^2)= S_4 - 2^3(1^2+2^2)= S_4 - 8S_2$. Now you can generalize this with $4,2$ are replaced by $2n, n$. Can you finish it? If the last term is odd, then isolate it and use up to the preceeding term which is even. So it can always be done this way...
A: $$\sum_{r=1}^{2n}(-1)^{r-1}r^2=\sum_{r=1}^n\{(2r-1)^2-(2r)^2\}=\sum_{r=1}^n(1-4r)$$
$$=\dfrac n2(1-4+1-4n)=\cdots=(-1)^{2n-1}\dfrac{2n(2n+1)}2$$
Now $$\sum_{r=1}^{2n+1}(-1)^{r-1}r^2=(2n+1)^2+\sum_{r=1}^{2n}(-1)^{r-1}r^2=?$$
A: For odd $n$, $(-1)^{n-1}=1$, so
$$1^2-2^2+3^2-4^2+\cdots+n^2\\
=1^2+(-2^2+3^2)+(-4^2+5^2)+\cdots+(-(n-1)^2+n^2)\\
=1+\color{blue}{(-2+3)}(2+3)+\color{blue}{(-4+5)}(4+5)+\cdots+\color{blue}{(-(n-1)+n)}((n-1)+n)\\
=\color{blue}{1\cdot}\left(1+2+3+4+5+\cdots+(n-1)+n\right)\\
=\frac {n(n+1)}2\\
=(-1)^{n-1}\frac {n(n+1)}2$$
For even $n$, $(-1)^{n-1}=-1$, so
$$1^2-2^2+3^2-4^2+\cdots+n^2\\
=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+\cdots+((n-1)^2-n^2)\\
=\color{orange}{(1-2)}(1+2)+\color{orange}{(3-4)}(3+4)+\color{orange}{(5-6)}(5+6)+\cdots+\color{orange}{((n-1)-n)}((n-1)+n)\\
=\color{orange}-(1+2+3+4+5+\cdots+(n-1)+n)\\
=-\frac {n(n+1)}2\\
=(-1)^{n-1}\frac {n(n+1)}2$$
Hence, for both odd and even $n$, 
$$1^2-2^2+3^2-4^2+\cdots+n^2=(-1)^{n-1}\frac {n(n+1)}2$$
A: For example, if $n$ is odd, $n=2k-1$ then we obtain:
$$LS=\frac{n(n+1)(2n+1)}{6}-8\cdot\frac{k(k-1)(2k-1)}{6}=$$
$$=\frac{n(n+1)(2n+1)}{6}-8\cdot\frac{\frac{n+1}{2}\cdot\frac{n-1}{2}\cdot n}{6}=\frac{n(n+1)}{2}$$
A: Note:  $(n+1)^2 = n^2 + 2n +1$.
So $(2n-1)^2 - (2n)^2 = 4n^2 -4n +1 - 4n^2 = -4n+1$.
So $\sum_{k=1}^{n=2m} (-1)^{k-1}k^2 = \sum_{j=1}^m [(2j-1)^2 - (2j)^2]=\sum_{j=1}^m [-4j +1]= -4(\sum j) + m = -4*\frac {m(m+1)}2 + m = -2m(m+1) +m=m(-2(m+1) + 1)=m(-2m-1)=-m(2m+1)= -\frac{n(n+1)}{2}$
..... if $n$ is even.
If $n=2m+1$ is odd then $\sum_{k=1}^{n=2m+1}(-1)^{k-1}k^2 = [\sum_{k=1}^{2m=2n-1}(-1)^{k-1}k^2] + n^2$
$= -\frac {(n-1)((n-1) + 1)}2 + n^2= n^2 - \frac {n(n-1)}2 = \frac {2n^2 -n(n-1)}2 = \frac {n(2n - (n-1)}2 = \frac {n(n+1)}2$
So either way ...
A: Here's a nice trick. For each $k$,
$$k^2=\frac{(k-1)k}2+\frac{k(k+1)}2=a_{k-1}+a_k$$
say.
Then
$$1^2-2^2+3^2-4^2+\cdots\pm n^2
=a_0+a_1-a_1-a_2+a_2+a_3-\cdots\pm a_{n-1}\mp a_n$$
and almost all of these terms cancel.
