Can we sum the first $2n$ terms of $1^2-3^2+5^2-7^2+\cdots$?

Can we sum the first $2n$ terms of $1^2-3^2+5^2-7^2+\cdots$?

Question

By bracketing the terms in pairs, or otherwise, find the sum of $2n$ terms of the series $1^2-3^2+5^2-7^2+\cdots.$

Hence show that the sum of $2n+1$ terms is $8n^2+8n+1$.

Suggestion

Can we find an expression for a general term in the series?

What is the sign of the $2k^{th}$ term?

Can we find expressions for the two terms in the $k^{th}$ pair?

Can we factorise $a^2-b^2$?

Can we check our formula by substituting in a couple of values for $n$?

Solution

We’ll simplify this sum by pairing the terms, as the question suggests. We know how to factorise the difference of two squares:

$$a^2-b^2=(a+b)(a-b).$$

We have here

$$\begin{align*} 1^2-3^2&=(1-3)(1+3)=-2\times 4, \\ 5^2-7^2&=(5-7)(5+7)=-2\times12, \\ 9^2-11^2&=(9-11)(9+11)=-2\times20, \\ \dots. \end{align*}$$

The $k^{th}$ pair of terms is $(4k-3)^2-(4k-1)^2=\bigl((4k-3)-(4k-1)\bigr)\bigl((4k-3)+(4k-1)\bigr) =-2(8k-4).$

How did we work out the $4k-3$ and $4k-1$?

The first terms of each pair form the sequence $1$, $5$, $9$, …, which is the sequence $4k-3$.

The second term of each pair is two greater, and so is $4k-1$.

So the sum of the first $2n$ terms, which is the first $n$ pairs, is $(-2)[4+12+20+... +(8n-4)]$ $=(-2)\dfrac{n}{2}[8+(n-1)8]$ $= -8n^2.$

We are using here that the sum of $n$ terms of an arithmetic sequence with first term $a$ and common difference $d$ is $\dfrac{n}{2}[2a+(n-1)d].$

We can check this: the sum of the first two terms is$1^2-3^2=-8=-8\times1^2$.

The sum of the first four terms is $1^2-3^2+5^2-7^2=-8+25-49=-32=-8\times 2^2$.

So it looks as if we have the correct formula.

Now we wish to find the sum of the first $2n+1$ terms.

The sum of the first $2n$ terms is $-8n^2$, and the $(2n+1)^{st}$ term is $(4(n+1)-3)^2=(4n+1)^2$.

Thus the sum of the first $2n+1$ terms is $(4n+1)^2-8n^2 = 8n^2+8n+1$, as required.