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

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

$$\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{\mathrm{5}} \:\mathrm{in}\:\left(\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +...\right)^{−\mathrm{3}/\mathrm{2}} \\ $$$$\mathrm{is}\:\mathrm{21}. \\ $$

Commented by gunawan last updated on 11/Jul/19

true or false

$$\mathrm{true}\:\mathrm{or}\:\mathrm{false} \\ $$

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

$${S}=\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} +... \\ $$$${Sx}=\:\:\:\:\:\:{x}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{4}} +... \\ $$$${S}−{Sx}=\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +... \\ $$$${S}\left(\mathrm{1}−{x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$${S}=\left(\mathrm{1}−{x}\right)^{−\mathrm{2}} \\ $$$$\left\{\left(\mathrm{1}−{x}\right)^{−\mathrm{2}} \right\}^{\frac{−\mathrm{3}}{\mathrm{2}}} \\ $$$$=\left(\mathrm{1}−{x}\right)^{\mathrm{3}} \\ $$$$=\mathrm{1}−\mathrm{3}{x}+\mathrm{3}{x}^{\mathrm{2}} −{x}^{\mathrm{3}} \\ $$$${so}\:{coefficient}\:{of}\:{x}^{\mathrm{5}} \:{is}\:{zero} \\ $$$${the}\:{given}\:{statement}\:{is}\:{false} \\ $$

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