If (1+x)^{15}=a_0+a_1 x+ldots+a_{15} x^{15},t | vedclass

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(C) Given that $(1+x)^{15} = \sum_{r=0}^{15} {}^{15}C_r x^r = a_0 + a_1 x + \ldots + a_{15} x^{15}$.Comparing coefficients,we have $a_r = {}^{15}C_r$.

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(C) Given that $(1+x)^{15} = \sum_{r=0}^{15} {}^{15}C_r x^r = a_0 + a_1 x + \ldots + a_{15} x^{15}$.Comparing coefficients,we have $a_r = {}^{15}C_r$.We need to evaluate $\sum_{r=1}^{15} r \frac{a_r}{a_{r-1}}$.Using the property $\frac{{}^{n}C_r}{{}^{n}C_{r-1}} = \frac{n-r+1}{r}$,we get $\frac{a_r}{a_{r-1}} = \frac{15-r+1}{r} = \frac{16-r}{r}$.Thus,$r \frac{a_r}{a_{r-1}} = r \cdot \frac{16-r}{r} = 16-r$.Therefore,$\sum_{r=1}^{15} (16-r) = (16-1) + (16-2) + \ldots + (16-15) = 15 + 14 + \ldots + 1$.This is the sum of the first $15$ natural numbers: $\frac{15(15+1)}{2} = \frac{15 	imes 16}{2} = 15 	imes 8 = 120$.

If $(1+x)^{15}=a_0+a_1 x+\ldots+a_{15} x^{15}$,then $\sum_{r=1}^{15} r \frac{a_r}{a_{r-1}}$ is equal to

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