Use induction to prove that sum of the first $2n$ terms of the series $1^2-3^2+5^2-7^2+…$ is $-8n^2$. Use induction to prove that sum of the first $2n$ terms of the series $1^2-3^2+5^2-7^2+…$ is $-8n^2$.
I came up with the formula $\displaystyle\sum_{r=1}^{2n} (-1)^{r+1}(2r-1)^2=-8n^2$ but I got stuck proving it by induction. Should I use another formula or is it correct and I should continue trying to prove it by induction?
 A: First, show that this is true for $n=1$:
$\sum\limits_{r=1}^{2}(-1)^{r+1}\cdot(2r-1)^2=-8$
Second, assume that this is true for $n$:
$\sum\limits_{r=1}^{2n}(-1)^{r+1}\cdot(2r-1)^2=-8n^2$
Third, prove that this is true for $n+1$:
$\sum\limits_{r=1}^{2(n+1)}(-1)^{r+1}\cdot(2r-1)^2=$
$\color{red}{\sum\limits_{r=1}^{2n}(-1)^{r+1}\cdot(2r-1)^2}+\color{green}{(-1)^{(2n+1)+1}\cdot(2(2n+1)-1)^2}+\color{blue}{(-1)^{(2n+2)+1}\cdot(2(2n+2)-1)^2}=$
$\color{red}{-8n^2}+\color{green}{(-1)^{(2n+1)+1}\cdot(2(2n+1)-1)^2}+\color{blue}{(-1)^{(2n+2)+1}\cdot(2(2n+2)-1)^2}=$
$-8n^2-8(2n+1)=$
$-8(n^2+2n+1)=$
$-8(n+1)^2$

Please note that the assumption is used only in the part marked red.
A: Just use the induction and get to
$$(4n+1)^2-(4n+3)^2-8n^2=-8(n+1)^2.$$
That is
$$(4n+1)^2-(4n+3)^2=-8(n+1)^2+8n^2=-8(2n+1).$$
Which is easy to verify.
A: For $n=1$ we have:
$$1^2 - 3^2 = -8 = -8\cdot1^2$$
Now assume that the statement holds for some n. Then we have a sum:
$$1^2-3^2+...+ (4n-3)^2 - (4n - 1)^2= -8n^2$$
Adding the next two terms to both sides we get:
\begin{align}
&1^2-3^2+...+ (4n-3)^2 - (4n - 1)^2 + (4n + 1)^2 - (4n + 3)^2 \\
&= -8n^2 + (4n + 1)^2 - (4n + 3)^2\\
&=-8n^2+16n^2+8n+1-16n^2-24n-9 \\
&=-8n^2-16n-8 \\
&=-8(n+1)^2
\end{align}
