Review question

Can we sum $1^2-2^2+3^2-4^2+\dotsb+ (2n-1)^2-(2n)^2$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9819

Solution

Prove that the sum of the first $n$ terms of the arithmetical progression $a+(a+d)+(a+2d)+...,$ is $\frac{1}{2}n\lbrace{2a+(n-1)d}\rbrace.$

We need the value of the sum $S=a+(a+d)+\dotsb+(a+(n-2)d)+(a+(n-1)d).$

Reversing the order of the terms, we have $S=(a+(n-1)d)+(a+(n-2)d)+\dotsb+(a+d)+a.$

We can now add these expressions together, giving
$2S=[a+(a+(n-1)d)]+[(a+d)+(a+(n-2)d)]+\dotsb+[(a+(n-2)d)+(a+d)],$ which can be rewritten as $2S=[2a+(n-1)d]+[2a+(n-1)d]+\dotsb+[2a+(n-1)d]+[2a+(n-1)d],$ where the expression $[2a+(n-1)d]$ occurs $n$ times.

This simplifies to $2S=n[2a+(n-1)d],$ and on dividing by two, we have $S=\frac{1}{2}n\lbrace2a+(n-1)d\rbrace$.

Hence, or otherwise, show that $1^2-2^2+3^2-4^2+\dotsb+ (2n-1)^2-(2n)^2=-n(2n+1).$

Using the difference of two squares, we can write \begin{align*} & 1^2-2^2+3^2-4^2+\dotsb+(2n-1)^2-(2n)^2 \\ = & (1^2-2^2)+(3^2-4^2)+\dotsb+ ((2n-1)^2-(2n^2))\\ = & (1-2)(1+2)+(3-4)(3+4)+\dotsb+ ((2n-1)-(2n))((2n-1)+2n))\\ = & -1[3+7+11+\dotsb+ (4n-1)]. \end{align*}

Now $3+7+11+\dotsb+ (4n-1)$ is an arithmetic series with $n$ terms and common difference $4$, so its sum is $\dfrac{n}{2}(6+(n-1)4)=n(2n+1)$.

Thus (remembering the $-1$), we have $1^2-2^2+3^2-4^2+\dotsb+(2n-1)^2-(2n)^2=-n(2n+1),$ as required.

Deduce the sums of the series

1. $1^2-2^2+3^3-4^2+\dotsb+(2n-1)^2-(2n)^2+(2n+1)^2$,
Using our formula from earlier, we have \begin{align*} &1^2-2^2+3^2-4^2+\dotsb+(2n-1)^2-(2n)^2+(2n+1)^2 \\ & \qquad = [1^2-2^2+3^2-4^2+\dotsb+(2n-1)^2-(2n)^2]+(2n+1)^2 \\ & \qquad = -n(2n+1)+(2n+1)^2 \\ & \qquad = (2n+1)[-n+2n+1] \\ & \qquad = (2n+1)(n+1). \end{align*}

1. $21^2-22^2+23^2-24^2+\dotsb+39^2-40^2$.

Let $S_{2n}=1^2-2^2+3^2-4^2+\dotsb+(2n-1)^2-(2n)^2$.

So using our formula from earlier, we need to find $S_{40}-S_{20} = -20(40+1)-(-10(20+1))= -820+210 = -610$.