If ^nC_r = C_r and 2 frac{C_1}{C_0} + 4 frac{ | vedclass

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(B) The general term of the series is $T_r = 2r \frac{C_r}{C_{r-1}}$.Using the property $\frac{C_r}{C_{r-1}} = \frac{n-r+1}{r}$,we get:$T_r = 2r \left

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$300$

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(B) The general term of the series is $T_r = 2r \frac{C_r}{C_{r-1}}$.Using the property $\frac{C_r}{C_{r-1}} = \frac{n-r+1}{r}$,we get:$T_r = 2r \left( \frac{n-r+1}{r} ight) = 2(n-r+1)$.The sum is $\sum_{r=1}^n T_r = \sum_{r=1}^n 2(n-r+1) = 2 \left[ n^2 - \frac{n(n+1)}{2} + n ight] = 2 \left[ n^2 - \frac{n^2+n}{2} + n ight] = 2 \left[ \frac{2n^2 - n^2 - n + 2n}{2} ight] = n(n+1)$.Given $n(n+1) = 650$,we have $n^2 + n - 650 = 0$.Solving for $n$,$(n+26)(n-25) = 0$,so $n = 25$.Therefore,$^nC_2 = ^{25}C_2 = \frac{25 	imes 24}{2} = 300$.

If $^nC_r = C_r$ and $2 \frac{C_1}{C_0} + 4 \frac{C_2}{C_1} + 6 \frac{C_3}{C_2} + \dots + 2n \frac{C_n}{C_{n-1}} = 650$,then $^nC_2 =$

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