Question

The sum of the product of the $2n$ numbers $\pm 1,\pm 2,\pm 3,...,\pm n$ taking two at a time is

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

Answer 1

Answer: (D)

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

The series will be like the following
$T_1=-1\times 1=-1$
$T_2=-2\times 2=-2^2$
$T_3=-3\times 3=-3^2$
:
:
$T_n=-n^2$
Hence the series will be
$-1-2^2-3^2..-n^2$
$=-(1+2^2+3^2+4^2+..n^2)$
Hence the general term will be
$T_n=-n^2$
Therefore the sum of the series upto nth term will be
$\sum T_n$
$=-\sum n^2$
$=-\left[\frac{n(n+1)(2n+1)}{6}\right]$