Question

Evaluate $\sum _{r=0}^{20}{r}^2\left({}^{20}C_r\right)$

Answer 1

Step 1. Solve the given expression.

$\sum _{r=0}^{20}{r}^2\left({}^{20}C_r\right)$

$=\sum _{r=1}^{20}{r}^2.\frac{20}{r}\left({}^{19}C_{r-1}\right)$

$=20\sum _{r=1}^{20}r\left({}^{19}C_{r-1}\right)$

Step 2. By adding and subtracting $1$, we get

$=\underset{r=1}{\overset{20}{20\sum }}\left(r-1+1\right).\left({}^{19}C_{r-1}\right)$

$=20\left[\sum _{r=1}^{20}\left(r-1\right).\left({}^{19}C_{r-1}\right)+\sum _{r=1}^{20}\left({}^{19}C_{r-1}\right)\right]$

$=20\left[\sum _{r=2}^{20}\left(r-1\right).\frac{19}{r-1}\left({}^{18}C_{r-2}\right)+\sum _{r=1}^{20}\left({}^{19}C_{r-1}\right)\right]$

$=20\left[\sum _{r=2}^{20}19\left({}^{18}C_{r-2}\right)+\sum _{r=1}^{20}\left({}^{19}C_{r-1}\right)\right]$

$=20\left[19{\left(2\right)}^{18}+2^{19}\right]$

$=20.2^{18}\left[19+2\right]$

$=20.21.2^{18}$

Hence, the solution of $\sum _{r=0}^{20}{r}^2\left({}^{20}C_r\right)$ is $\mathbf{20}\mathbf{}\mathbf{.}\mathbf{}\mathbf{21}\mathbf{}\mathbf{.}\mathbf{}{\mathbf{2}}^{\mathbf{18}}$