The coefficient of x^{10} in the expansion of | vedclass

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Question

(C) The given series is a geometric progression with $a=1$,common ratio $r=(1+x)$,and $n=21$ terms.The sum of the series is $S = \frac{1((1+x)^{21}-1)

Vedclass · Accepted Answer

${ }^{21} C_{11}$

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(C) The given series is a geometric progression with $a=1$,common ratio $r=(1+x)$,and $n=21$ terms.The sum of the series is $S = \frac{1((1+x)^{21}-1)}{(1+x)-1} = \frac{(1+x)^{21}-1}{x}$.We need to find the coefficient of $x^{10}$ in $S = \frac{(1+x)^{21}-1}{x}$.This is equivalent to finding the coefficient of $x^{11}$ in the expansion of $(1+x)^{21}-1$.The general term in the expansion of $(1+x)^{21}$ is given by $T_{r+1} = {}^{21}C_{r} x^{r}$.For $r=11$,the term is ${}^{21}C_{11} x^{11}$.Thus,the coefficient of $x^{11}$ in $(1+x)^{21}$ is ${}^{21}C_{11}$.Therefore,the coefficient of $x^{10}$ in the given series is ${}^{21}C_{11}$.

The coefficient of $x^{10}$ in the expansion of $1+(1+x)+\dots+(1+x)^{20}$

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