Question

The sum of products of numbers $\pm 1,\pm 2,...\pm n,$taken two at a time is
$\begin{align}
  & A.\text{ }\dfrac{\text{n(n+1)}}{2} \\
 & B.\text{ }\dfrac{n(n+1)(2n+1)}{6} \\
 & C.\text{ }\dfrac{\text{-n(n+1)(2n+1)}}{6} \\
 & D.\text{ 0} \\
\end{align}$

Answer 1

Hint: First find the series by multiplication of numbers taken two at a time and then apply the formula of sum of the square of first n natural numbers.

Complete step-by-step answer:
we have sequence
$\pm 1,\pm 2,....\pm n$
So we form series taken two at a time let series be so
$\begin{align}
  & {{S}_{n}}=(-1)(+1)+(-2)(+2)+...+(-n)(+n) \\
 & \Rightarrow {{S}_{n}}=-{{1}^{2}}-{{2}^{2}}-{{3}^{2}}-...-{{n}^{2}} \\
\end{align}$

Taken negative sign outside
$\Rightarrow {{S}_{n}}=-({{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{n}^{2}})$
As we know
${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{n}^{2}}=\dfrac{n(n+1)(2n+1)}{6}$
So
$\Rightarrow {{S}_{n}}=-\dfrac{n(n+1)(2n+1)}{6}$

Hence option C is correct.

Note : In order to form a series of a given sequence, we put a ‘+’ sign between the sequence elements. Also we can use identity ${{k}^{3}}-{{(k-1)}^{3}}=3{{k}^{2}}-3k+1$ , so put value of k from 1 to n and after addition we can find the sum of square first n natural numbers.