If {}^{20}C_{r} is the coefficient of x^{r} i | vedclass

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(D) We know that $\sum_{r=0}^{n} r \cdot {}^{n}C_{r} = n \cdot 2^{n-1}$ and $\sum_{r=0}^{n} r(r-1) \cdot {}^{n}C_{r} = n(n-1) \cdot 2^{n-2}$.We can wr

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$420 	imes 2^{18}$

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(D) We know that $\sum_{r=0}^{n} r \cdot {}^{n}C_{r} = n \cdot 2^{n-1}$ and $\sum_{r=0}^{n} r(r-1) \cdot {}^{n}C_{r} = n(n-1) \cdot 2^{n-2}$.We can write $r^{2} = r(r-1) + r$.Therefore,$\sum_{r=0}^{20} r^{2} \cdot {}^{20}C_{r} = \sum_{r=0}^{20} [r(r-1) + r] \cdot {}^{20}C_{r}$.$= \sum_{r=0}^{20} r(r-1) \cdot {}^{20}C_{r} + \sum_{r=0}^{20} r \cdot {}^{20}C_{r}$.Using the identities with $n=20$:$= 20 	imes 19 	imes 2^{20-2} + 20 	imes 2^{20-1}$.$= 380 	imes 2^{18} + 20 	imes 2^{19}$.$= 380 	imes 2^{18} + 20 	imes 2 	imes 2^{18}$.$= 380 	imes 2^{18} + 40 	imes 2^{18}$.$= (380 + 40) 	imes 2^{18} = 420 	imes 2^{18}$.

If ${}^{20}C_{r}$ is the coefficient of $x^{r}$ in the expansion of $(1+x)^{20}$,then the value of $\sum_{r=0}^{20} r^{2} \cdot {}^{20}C_{r}$ is equal to:

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