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

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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 \[(4k-3)^2-(4k-1)^2=\bigl((4k-3)-(4k-1)\bigr)\bigl((4k-3)+(4k-1)\bigr) =-2(8k-4).\]

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 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.