If c_{0}, c_{1}, c_{2}, ldots, c_{15} are the | vedclass

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(B) The general term of the given series is $T_{r} = r \frac{c_{r}}{c_{r-1}}$.We know that $c_{r} = {}^{15}C_{r} = \frac{15!}{r!(15-r)!}$ and $c_{r-1}

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$120$

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(B) The general term of the given series is $T_{r} = r \frac{c_{r}}{c_{r-1}}$.We know that $c_{r} = {}^{15}C_{r} = \frac{15!}{r!(15-r)!}$ and $c_{r-1} = {}^{15}C_{r-1} = \frac{15!}{(r-1)!(16-r)!}$.Therefore, $\frac{c_{r}}{c_{r-1}} = \frac{15-r+1}{r} = \frac{16-r}{r}$.Substituting this into the general term, we get $T_{r} = r 	imes \frac{16-r}{r} = 16-r$.The sum is $S = \sum_{r=1}^{15} (16-r) = (16-1) + (16-2) + \ldots + (16-15) = 15 + 14 + \ldots + 1$.This is the sum of the first $15$ natural numbers, which is $\frac{n(n+1)}{2} = \frac{15 	imes 16}{2} = 120$.

If $c_{0}, c_{1}, c_{2}, \ldots, c_{15}$ are the binomial coefficients in the expansion of $(1+x)^{15}$, then the value of $\frac{c_{1}}{c_{0}}+2 \frac{c_{2}}{c_{1}}+3 \frac{c_{3}}{c_{2}}+\ldots+15 \frac{c_{15}}{c_{14}}$ is

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