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Binomial Theorem for Any Index

For the given...

Question

For the given expansion $(1 + 2 x + 3 x^{2} + . . . .)^{- 3 / 2}, (| x | < 1)$

sum of all coefficient will be

0

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coefficient of

x^{5}

will be

0

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sum of all coefficient will be

8

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coefficient of

x^{2}

will be

3

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Solution

The correct option is D coefficient of $x^{2}$ will be $3$
As we know that
$(1 - x)^{- 2} = 1 + 2 x + \frac{2 \cdot 3}{2} x^{2} + \frac{2 \cdot 3 \cdot 4}{2 \cdot 3} x^{3} + \dots = 1 + 2 x + 3 x^{2} + 4 x^{3} + \dots$

$∴ (1 + 2 x + 3 x^{2} + . . . .)^{- 3 / 2} = [(1 - x)^{- 2}]^{- 3 / 2}$
$= (1 - x)^{3} = 1 - 3 x + 3 x^{2} - x^{3}$

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If |x| < 1, then in the expansion of

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