Use induction to prove that sum of the first $2n$ terms of the series $1^2-3^2+5^2-7^2+…$ is $-8n^2$.

Question

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?

Answer 1

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 $n+1$:

$\sum\limits_{r=1}^{2(n+1)}(-1)^{r+1}\cdot(2r-1)^2=$

$\color{red}{\sum\limits_{r=1}^{2n}(-1)^{r+1}\cdot(2r-1)^2}+\color{green}{(-1)^{(2n+1)+1}\cdot(2(2n+1)-1)^2}+\color{blue}{(-1)^{(2n+2)+1}\cdot(2(2n+2)-1)^2}=$

$\color{red}{-8n^2}+\color{green}{(-1)^{(2n+1)+1}\cdot(2(2n+1)-1)^2}+\color{blue}{(-1)^{(2n+2)+1}\cdot(2(2n+2)-1)^2}=$

$-8n^2-8(2n+1)=$

$-8(n^2+2n+1)=$

$-8(n+1)^2$

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Please note that the assumption is used only in the part marked red.

Answer 2

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.

Answer 3

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}