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

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.

Hence show that the sum of \(2n+1\) terms is \(8n^2+8n+1\).

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.