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Mathematics Question on Sequence and series

The coefficient of x3x^3 in the infinite series expansion of 2(1−x)(2−x),\frac{2}{\left(1-x\right)\left(2-x\right)}, for ∣x∣<1,\left|x\right|< 1, is

A

−1/16-1/16

B

15/815/8

C

−1/8-1/8

D

15/1615/16

Answer

15/815/8

Explanation

Solution

Given expression can be rewritten as 2(1−x)−1×2−1(1−x2)−12(1-x)^{-1} \times 2^{-1}\left(1-\frac{x}{2}\right)^{-1} =[1+x+x2+x3+…]=\left[1+x+x^{2}+x^{3}+\ldots\right] [1+(x2)+(x2)2+(x2)3+…]\left[1+\left(\frac{x}{2}\right)+\left(\frac{x}{2}\right)^{2}+\left(\frac{x}{2}\right)^{3}+\ldots\right] Coefficient of x3x^{3} in the above expression is =[(12)3+(12)2+(12)+1]=\left[\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)+1\right] =[18+14+12+1]=\left[\frac{1}{8}+\frac{1}{4}+\frac{1}{2}+1\right] =[1+2+4+88]=158=\left[\frac{1+2+4+8}{8}\right]=\frac{15}{8}