Proving that the sum of the first $2n$ terms of the series $1^2 - 3^2 + 5^2 - \cdots$ is $-8n^2$ by induction

Question

Use mathematical induction to prove the following for the first $2n$ terms of the series $$1^2 - 3^2 + 5^2 - 7^2 + \cdots = -8n^2.$$

As we have odd numbers that are squared we could use $n = 2k-1$.

But the $2$ sides do not equate for $n=1$ or $2k-1$ (if you set $k=1$).

Also need to find the sum to $2n+1$ terms.

Answer 1

Here is the main part of the inductive step: \begin{align} \sum_{i=1}^{2(k+1)}(-1)^{i-1}(2i-1)^2&=\sum_{i=1}^{2k}(-1)^{i-1}(2i-1)^2+\underbrace{(-1)^{2k+1}(4k+3)^2}_{i=2k+2}+\underbrace{(-1)^{2k}(4k+1)^2}_{i=2k+1}\\[0.5em] &= -8k^2-(4k+3)^2+(4k+1)^2\\[0.5em] &= -8k^2-16k-8\\[0.5em] &= -8(k^2+2k+1)\\[0.5em] &= -8(k+1)^2. \end{align} Do you see where the induction hypothesis was used?

Answer 2

As we have odd numbers that are squared we could use n = 2k-1

You are summing the sequence: $\{1^2-3^2, 5^2-7^2, 9^2-11^2, \ldots (4n-3)^2-(4n-1)^2, \ldots\}$

But the 2 sides do not equate for n=1 or 2k-1 (if you set k=1)

See above. The first term is $-8$, and so...

Also need to find the sum to (2n+1) terms.

You need to show that if the sum of the first $n$ terms is $-8 n^2$, then if you add $-16(n+1)+8$ you obtain $-8(n+1)^2$.

Answer 3

$$a_n = \sum_{k=1}^{n}(-1)^{k+1}(2k-1)^2 $$

You need to rewrite your function to accommodate your analysis. If you have to analyse every pair of values you could rewrite it as:

$$b_m = \sum_{k=1}^{m}(4k-3)^2-(4k-1)^2$$

But that's

$$\sum_{k=1}^{m}(4k-3)^2-(4k-1)^2 = \sum_{k=1}^{m}(16k^2-24k+9 - 16k^2+8k-1) = \sum_{k=1}^{m}(-16k+8)$$

So:

for $$n = 2 \implies m = 1$$

$$\sum_{k=1}^{1}(-16k+8) = -16 + 8 = -8$$

Now, assuming for $2n$ works, what's $2n+2$ in $a_n$. But that's the same for testing $m+1$ assuming $m$ is true for $b_n$

$$\sum_{k=1}^{m+1}(-16k+8) = \left(\sum_{k=1}^{m}(-16k+8)\right) -16(m+1)+8 = -8m^2 -16m -16 +8 =-8m^2 -16m -8 = -8(m+1)^2$$

Answer 4

$1^2−3^2+5^2−7^2+⋯=−8n^2$

Last term of this series is $(4n-3)^2-(4n-1)^2$

$\Rightarrow P(n):1^2−3^2+5^2−7^2+⋯+(4n-3)^2-(4n-1)^2=−8n^2$

For n=1 $\Rightarrow P(1):1^2−3^2=−8(1)^2$

$\Rightarrow P(1):-8=−8$ which is true

Let P(k)be true, $P(k):1^2−3^2+5^2−7^2+⋯+(4k-3)^2-(4k-1)^2=−8k^2$.......(1)

We have to prove P(k+1) as true.

$P(k+1):1^2−3^2+5^2−7^2+⋯+(4k-3)^2-(4k-1)^2+(4(k+1)-3)^2-(4(k+1)-1)^2=−8(k+1)^2$

$LHS=1^2−3^2+5^2−7^2+⋯+(4k-3)^2-(4k-1)^2+(4k+1)^2-(4k+3)^2$

$=-8k^2+(4k+1)^2-(4k+3)^2$ $=-8k^2+16k^2+8k+1-(16k^2+24k+9)$
$=-8k^2-16k-8=-8(k^2+2k+1)=-8(k+1)^2$=RHS

Therefore P(k+1) is true.So by Induction principle,P(n) is true for all $n$