If the sum of the coefficients of x^r (r=0, 1 | vedclass

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(D) The sum of the coefficients in the expansion of $P(x) = (1+3x-2x^2)^n$ is obtained by setting $x=1$. Given $P(1) = (1+3(1)-2(1)^2)^n = 128$. $(1+3

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$105$

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(D) The sum of the coefficients in the expansion of $P(x) = (1+3x-2x^2)^n$ is obtained by setting $x=1$. Given $P(1) = (1+3(1)-2(1)^2)^n = 128$. $(1+3-2)^n = 128$ $\Rightarrow 2^n = 2^7$ $\Rightarrow n=7$. We need to evaluate the sum $S = \sum_{r=1}^{2n} r \frac{^{2n}C_r}{^{2n}C_{r-1}}$. Using the property $\frac{^{n}C_r}{^{n}C_{r-1}} = \frac{n-r+1}{r}$,we have: $\frac{^{2n}C_r}{^{2n}C_{r-1}} = \frac{2n-r+1}{r}$. Substituting this into the sum: $S = \sum_{r=1}^{2n} r \left( \frac{2n-r+1}{r} ight) = \sum_{r=1}^{2n} (2n-r+1)$. This is an arithmetic progression with $2n$ terms. $S = (2n) + (2n-1) + \ldots + 1 = \frac{(2n)(2n+1)}{2} = n(2n+1)$. For $n=7$,$S = 7(2(7)+1) = 7 	imes 15 = 105$.

If the sum of the coefficients of $x^r$ $(r=0, 1, 2, \ldots, 2n)$ in the expansion of $(1+3x-2x^2)^n$ is $128$,then $\sum_{r=1}^{2n} r \frac{^{2n}C_r}{^{2n}C_{r-1}} = $

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