Question

`Find } \sum r^(2) \cdot{ }^(20) C_(r)\end{equation}`

Options:
(a)
`$2^(20)$`
(b)
`$2^(21)$`
(c)
`$210 * 2^(19)$`

Answer 1

I bet the sum you're referring to is supposed to be

`\displaystyle \sum_(r=0)^(20) r^2 * {}^(20)C_r`

or equivalently,

`\displaystyle \sum_(r=0)^(20) r^2 \binom{20}r`

where
`\binom nk = (n!)/(k!(n-k)!)` is the binomial coefficient.

Recall the binomial series,

`(1+x)^\alpha = \displaystyle \sum_(r=0)^\infty \binom\alpha r x^r`

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

`\alpha (1+x)^(\alpha-1) = \displaystyle \sum_(r=0)^\infty r \binom\alpha r x^(r-1)`

Multiply both sides by some arbitrary x in |x| < 1 :

`\alpha x (1+x)^(\alpha-1) = \displaystyle \sum_(r=0)^\infty r \binom\alpha r x^r`

Repeat:

`\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)`

`\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`

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

`20 * 2^(19) + 20*19* 2^(18) = (20 + 10*19)*2^(19) = \boxed{210*2^(19)}`