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 \(S_{40}-S_{20} = -20(40+1)-(-10(20+1))= -820+210 = -610\).