Question

The sum of the series$1+2\times 3+3\times 5+4\times 7+....$ upto $11^{th}$ term is :-

• A. 915
• B. 946
• C. 945
• D. 916

Answer 1

Answer: (B)

$T_r=r\left(2r-1\right)$
$S=\sum 2r^2-\sum r$
$S=\frac{2.n\left(n+1\right)\left(2n+1\right)}{6}-\frac{n\left(n+1\right)}{2}$
$S_{11}=\frac{2}{6}.\left(11\right)\left(12\right)\left(23\right)-\frac{11\left(12\right)}{2}=\left(44\right)\left(23\right)-66=946$