The coefficient of $x^{8}$ in the expansion of ${(1 + 2 x + 3 x^{2} + 4 x^{3} + . . . .)}^{\frac{1}{2}}$ is

1

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0

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None of these

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Solution

The correct option is A $1$
Simplifying the above expression, we get
$S = {(\frac{1}{(1 - x)^{2}})}^{\frac{1}{2}}$
$= {((1 - x)^{- 2})}^{\frac{1}{2}}$
$= (1 - x)^{- 1}$
Hence
$S = 1 + x + x^{2} + x^{3} + . . . \infty$
Hence the coefficient of $x^{8}$ in S is 1.

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