The coefficient of x^5 in (1+2x+3x^2 +...)^(−3/2) is 21.


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Question Number 63907 by gunawan last updated on 11/Jul/19

$The coefficient of x^{5} in {(1 + 2 x + 3 x^{2} + . . .)}^{- 3 / 2}$ $is 21 .$
Commented by gunawan last updated on 11/Jul/19

$true or false$
	Answered by Hope last updated on 11/Jul/19
	$S=1+2x+3x^2 +4x^3 +... Sx= x+2x^2 +3x^3 +4x^4 +... S−Sx=1+x+x^2 +x^3 +... S(1−x)=(1/(1−x)) S=(1−x)^(−2) {(1−x)^(−2) }^((−3)/2) =(1−x)^3 =1−3x+3x^2 −x^3 so coefficient of x^5 is zero the given statement is false$
	$S = 1 + 2 x + 3 x^{2} + 4 x^{3} + . . .$ $S x = x + 2 x^{2} + 3 x^{3} + 4 x^{4} + . . .$ $S - S x = 1 + x + x^{2} + x^{3} + . . .$ $S (1 - x) = \frac{1}{1 - x}$ $S = {(1 - x)}^{- 2}$ ${{(1 - x)}^{- 2}}^{\frac{- 3}{2}}$ $= {(1 - x)}^{3}$ $= 1 - 3 x + 3 x^{2} - x^{3}$ $s o c o e f f i c i e n t o f x^{5} i s z e r o$ $t h e g i v e n s t a t e m e n t i s f a l s e$
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