Question

If $S_n=1^2+{2.2}^2+3^2+{2.4}^2+....=\frac{n(n+1{)}^2}{2}$ where $n$ is even, then the value of $S_{21}$ must be

• A. $(11{)}^2\times 21$
• B. $11\times (21{)}^2$
• C. $22\times (21{)}^2$
• D. None of these

Answer 1

The correct option is B $11\times (21{)}^2$

Given $S_n=1^2+{2.2}^2+3^2+{2.4}^2+....=\frac{n(n+1{)}^2}{2}$ for $n$ is even

$S_{21}=S_{20}+(21{)}^2$

$=\frac{20\times (21{)}^2}{2}+441$

$=\left(21{)}^2\right(11)=441\times 11=4851$

Hence choice (b) is correct answer.