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  4. The coefficient of x5 in (1 + 2x + 3x2 + ....)-7/2 is

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Q. The coefficient of $x^5$ in $(1 + 2x + 3x^2 + ....)^{-7/2}$ is

Binomial Theorem

Solution:

$1 + 2x + 3x^{2} + .... = \left(1 + x\right) ^{- 2}$
$\Rightarrow \left(1+2x+3x^{2} +.... \right)^{-3/ 2} = \left\{\left(1+x\right)^{-2}\right\}^{-7/ 2} = \left(1+x\right)^{7}$
$\therefore \quad$ The coefficient of $x^{5}$ in $\left(1 + 2x + 3x^{2} + ........\right)^{- 7/2}$
= Coefficient of $x^{5}$ in $\left(1 + x\right)^{7}$
$= ^{7}C_{5} = 21$