The coefficient of x^{50} in the expansion of | vedclass

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(C) Let $S = \sum_{k=1}^{101} k x^{k-1} (1+x)^{101-k}$.This is an arithmetico-geometric series of the form $\sum_{k=1}^{n} k a^{n-k} b^{k-1}$.Let $a =

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$^{102}C_{50}$

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(C) Let $S = \sum_{k=1}^{101} k x^{k-1} (1+x)^{101-k}$.This is an arithmetico-geometric series of the form $\sum_{k=1}^{n} k a^{n-k} b^{k-1}$.Let $a = (1+x)$ and $b = x$. Then $S = \sum_{k=1}^{101} k (1+x)^{101-k} x^{k-1}$.Multiplying by $\frac{x}{1+x}$,we get $\frac{x}{1+x} S = \sum_{k=1}^{101} k (1+x)^{100-k} x^k$.Subtracting the two expressions:$S(1 - \frac{x}{1+x}) = (1+x)^{100} + x(1+x)^{99} + x^2(1+x)^{98} + \dots + x^{100} - 101 x^{100} \cdot \frac{x}{1+x}$.$S(\frac{1}{1+x}) = \frac{(1+x)^{101} - x^{101}}{(1+x) - x} - 101 \frac{x^{101}}{1+x} = (1+x)^{101} - x^{101} - 101 \frac{x^{101}}{1+x}$.$S = (1+x)^{102} - x^{101}(1+x) - 101 x^{101} = (1+x)^{102} - x^{101} - x^{102} - 101 x^{101} = (1+x)^{102} - x^{102} - 102 x^{101}$.The coefficient of $x^{50}$ in $(1+x)^{102}$ is $^{102}C_{50}$.Since $x^{102}$ and $x^{101}$ do not contain $x^{50}$,the coefficient is $^{102}C_{50}$.

The coefficient of $x^{50}$ in the expansion of $(1+x)^{100}+2x(1+x)^{99}+3x^2(1+x)^{98}+\dots+101x^{100}$ is:

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