The coefficient of x^{91} in the series ^{100 | vedclass

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(C) The given series is $S = \sum_{r=1}^{9} {^{100}C_r \cdot 2^{9-r} \cdot (1-x)^{100-r}}$.We need the coefficient of $x^{91}$ in this expression.The

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$^{100}C_{9}(2^9 - 3^9)$

Vedclass · Answer

(C) The given series is $S = \sum_{r=1}^{9} {^{100}C_r \cdot 2^{9-r} \cdot (1-x)^{100-r}}$.We need the coefficient of $x^{91}$ in this expression.The term $(1-x)^{100-r}$ can be expanded as $\sum_{k=0}^{100-r} {^{100-r}C_k (-x)^k}$.The coefficient of $x^{91}$ in $(1-x)^{100-r}$ is $^{100-r}C_{91} (-1)^{91} = -^{100-r}C_{91}$.Thus,the coefficient of $x^{91}$ in the series is $\sum_{r=1}^{9} {^{100}C_r \cdot 2^{9-r} \cdot (-^{100-r}C_{91})}$.Using the identity $^{n}C_r \cdot ^{n-r}C_k = ^{n}C_k \cdot ^{n-k}C_r$,we have $^{100}C_r \cdot ^{100-r}C_{91} = ^{100}C_{91} \cdot ^{100-91}C_r = ^{100}C_9 \cdot ^{9}C_r$.Substituting this,the coefficient is $-\sum_{r=1}^{9} {^{100}C_9 \cdot ^{9}C_r \cdot 2^{9-r}} = -^{100}C_9 \sum_{r=1}^{9} {^{9}C_r \cdot 2^{9-r}}$.Using the Binomial Theorem,$(2+1)^9 = \sum_{r=0}^{9} {^{9}C_r \cdot 2^{9-r} \cdot 1^r} = ^{9}C_0 \cdot 2^9 + \sum_{r=1}^{9} {^{9}C_r \cdot 2^{9-r}}$.So,$\sum_{r=1}^{9} {^{9}C_r \cdot 2^{9-r}} = 3^9 - 2^9$.Therefore,the coefficient is $-^{100}C_9 (3^9 - 2^9) = ^{100}C_9 (2^9 - 3^9)$.

The coefficient of $x^{91}$ in the series $^{100}C_1 \cdot 2^8 \cdot (1 - x)^{99} + ^{100}C_2 \cdot 2^7 \cdot (1 - x)^{98} + ^{100}C_3 \cdot 2^6 \cdot (1 - x)^{97} + \dots + ^{100}C_9 \cdot (1 - x)^{91}$ is equal to -

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