I bet the sum you're referring to is supposed to be
[tex]\displaystyle \sum_{r=0}^{20} r^2 \times {}^{20}C_r[/tex]
or equivalently,
[tex]\displaystyle \sum_{r=0}^{20} r^2 \binom{20}r[/tex]
where [tex]\binom nk = \frac{n!}{k!(n-k)!}[/tex] is the binomial coefficient.
Recall the binomial series,
[tex](1+x)^\alpha = \displaystyle \sum_{r=0}^\infty \binom\alpha r x^r[/tex]
which is valid for |x| < 1. (Note that if r > α, the binomial coefficient is defined to be zero, so there really are only α many terms when α is a whole number.)
Differentiating both sides with respect to x gives
[tex]\alpha (1+x)^{\alpha-1} = \displaystyle \sum_{r=0}^\infty r \binom\alpha r x^{r-1}[/tex]
Multiply both sides by some arbitrary x in |x| < 1 :
[tex]\alpha x (1+x)^{\alpha-1} = \displaystyle \sum_{r=0}^\infty r \binom\alpha r x^r[/tex]
Repeat:
[tex]\alpha (1+x)^{\alpha-1} + \alpha(\alpha-1) x(1+x)^{\alpha-2} = \displaystyle \sum_{r=0}^\infty r^2 \binom\alpha r x^{r-1}[/tex]
[tex]\alpha x (1+x)^{\alpha-1} + \alpha(\alpha-1) x^2 (1+x)^{\alpha-2} = \displaystyle \sum_{r=0}^\infty r^2 \binom\alpha r x^r[/tex]
Let α = 20, and let x approach 1 from below. The right side converges to the sum we want, while the left side converges to
[tex]20 \times 2^{19} + 20\times19\times 2^{18} = (20 + 10\times19)\times2^{19} = \boxed{210\times2^{19}}[/tex]
