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

Given series is $1+2\times 3+3\times 5+4\times 7+5\times 9+\dots$

Thus, the general term is $T_r=r\left(2r-1\right)$

Hence, the sum of $11$ terms of the series is $S_{11}=\sum _{r=1}^{11}r\left(2r-1\right)$

$\Rightarrow S_{11}=2\sum r^2-\sum r$

Using the formulae for $\underset{r=1}{\overset{n}{\sum r^2}}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$ and $\underset{r=1}{\overset{n}{\sum r}}=\frac{n\left(n+1\right)}{2},$ we get

$S_{11}=2\cdot \frac{11\left(11+1\right)\left(22+1\right)}{6}-\frac{11\left(11+1\right)}{2}$

$\Rightarrow S_{11}=44\times 23-66$

$\Rightarrow S_{11}=1012-66=946.$