Question

The sum of the products of the numbers $\pm 1,\pm 2,...\pm n$, taken two at a time is

• A. $\frac{n(n+1)}{2}$
• B. $\frac{n(n+1)(2n+1)}{6}$
• C. $\frac{-n(n+1)(2n+1)}{6}$
• D. $0$

Answer 1

Answer: (C)

$\frac{-n(n+1)(2n+1)}{6}$

${[\pm 1,\pm 2,\pm 3,.....\pm n]}^2$ is equal to
$[(1+2+.....+n)+(-1)+(-2)+....+{(-n)}^2]=2[1^2+2^2+....+n^2]+2[Producttakentwoatatime]$
(Apply $[{(a+b+c+d+...)}^2=(a^2+b^2+c^2+....)+2(ab+bc+ac+ad+......)]$)
$\Rightarrow 0=2[1^2+2^2+....+n^2]+2\left[Prodcuttakentwoatatime\right]$
$=2\left[Prodcuttakentwoatatime\right]=-2[1^2+2^2+....+n^2]$
$=-2\left[\frac{n(n+1)(2n+1)}{6}\right]=-\left(\frac{n(n+1)(2n+1)}{6}\right)$