frac{C_1}{C_0} + 2frac{C_2}{C_1} + 3frac{C_3} | vedclass

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(B) The general term of the series is given by $k \frac{C_k}{C_{k-1}}$.We know that $C_k = \binom{n}{k} = \frac{n!}{k!(n-k)!}$.Thus,$\frac{C_k}{C_{k-1

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(B) The general term of the series is given by $k \frac{C_k}{C_{k-1}}$.We know that $C_k = \binom{n}{k} = \frac{n!}{k!(n-k)!}$.Thus,$\frac{C_k}{C_{k-1}} = \frac{n! / (k!(n-k)!)}{n! / ((k-1)!(n-k+1)!)} = \frac{(k-1)!(n-k+1)!}{k!(n-k)!} = \frac{n-k+1}{k}$.Therefore,the $k$-th term is $k 	imes \frac{n-k+1}{k} = n-k+1$.The sum is $\sum_{k=1}^{n} (n-k+1) = n + (n-1) + \dots + 1 = \frac{n(n+1)}{2}$.For $n=15$,the sum is $\frac{15(15+1)}{2} = \frac{15 	imes 16}{2} = 15 	imes 8 = 120$.

$\frac{C_1}{C_0} + 2\frac{C_2}{C_1} + 3\frac{C_3}{C_2} + \dots + 15\frac{C_{15}}{C_{14}} = $

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