In the expansion of (1+x)^n,the sum frac{C_1} | vedclass

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

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$\frac{n(n+1)}{2}$

Vedclass · Answer

(A) We know that $C_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$.Thus,$\frac{C_r}{C_{r-1}} = \frac{n!}{r!(n-r)!} 	imes \frac{(r-1)!(n-r+1)!}{n!} = \frac{n-r+1}{r}$.The given expression is $S = \sum_{r=1}^{n} r \frac{C_r}{C_{r-1}}$.Substituting the ratio: $S = \sum_{r=1}^{n} r \left( \frac{n-r+1}{r} ight) = \sum_{r=1}^{n} (n-r+1)$.Expanding the sum: $S = n + (n-1) + (n-2) + \ldots + 1$.This is the sum of the first $n$ natural numbers,which is $\frac{n(n+1)}{2}$.

In the expansion of $(1+x)^n$,the sum $\frac{C_1}{C_0} + 2 \frac{C_2}{C_1} + 3 \frac{C_3}{C_2} + \ldots + n \frac{C_n}{C_{n-1}}$ is equal to:

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