S = 1 + 2 × 3 + 3 × 5 + 4 × 7 + ....... + upto 11 terms

n^th term of the series is T_n = n(2n–1)

$\Rightarrow \mathrm{S}=\sum _{\mathrm{n}=1}^{11}{\mathrm{T}}_{\mathrm{n}}=\sum _{\mathrm{n}=1}^{11}(2{\mathrm{n}}^2-\mathrm{n})$

$\Rightarrow {\mathrm{S}}_{\mathrm{n}}=\frac{2\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}-\frac{\mathrm{n}(\mathrm{n}+1)}{2}$

Put n = 1

$\Rightarrow {\mathrm{S}}_{11}=\frac{2\left(11\right)\left(12\right)\left(23\right)}{6}-\frac{11\left(12\right)}{2}$

⇒ S₁₁ = 946