If C_0, C_1, C_2, ldots, C_8 are the binomial | vedclass

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(A) We know that the binomial coefficient $C_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$.Thus,$\frac{C_r}{C_{r-1}} = \frac{n!}{r!(n-r)!} 	imes \frac{(r-1

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$540$

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(A) We know that the binomial coefficient $C_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$.Thus,$\frac{C_r}{C_{r-1}} = \frac{n!}{r!(n-r)!} 	imes \frac{(r-1)!(n-r+1)!}{n!} = \frac{n-r+1}{r}$.Given $n=8$,we have $\frac{C_r}{C_{r-1}} = \frac{8-r+1}{r} = \frac{9-r}{r}$.Now,the sum is $S = \sum_{r=1}^8 r^3 \left( \frac{9-r}{r} ight) = \sum_{r=1}^8 r^2(9-r) = \sum_{r=1}^8 (9r^2 - r^3)$.Using the summation formulas $\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{r=1}^n r^3 = \left( \frac{n(n+1)}{2} ight)^2$ for $n=8$:$\sum_{r=1}^8 r^2 = \frac{8 	imes 9 	imes 17}{6} = 204$.$\sum_{r=1}^8 r^3 = \left( \frac{8 	imes 9}{2} ight)^2 = 36^2 = 1296$.Therefore,$S = 9(204) - 1296 = 1836 - 1296 = 540$.

If $C_0, C_1, C_2, \ldots, C_8$ are the binomial coefficients in the expansion of $(1+x)^8$,then $\sum_{r=1}^8 r^3 \frac{C_r}{C_{r-1}} =$

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