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The coefficient of x^n in the expansion of \left(\frac{1}{1-x}\right)\left(\frac{1}{3-x}\right) \text { is }

Option: 1

\frac{3^{n+1}-1}{2 \cdot 3^{n+1}}


Option: 2

\frac{3^{n+1}-1}{3^{n+1}}


Option: 3

2\left(\frac{3^{n+1}-1}{3^{n+1}}\right)


Option: 4

None of these


Answers (1)

best_answer

\begin{gathered} \frac{1}{(1-x)(3-x)}=(1-x)^{-1}(3-x)^{-1}=3^{-1}(1-x)^{-1}\left(1-\frac{x}{3}\right)^{-1}= \\ \frac{1}{3}\left[1+x+x^2+\ldots . . x^n\right]\left[1+\frac{x}{3}+\frac{x^2}{3^2}+\ldots . .+\frac{x^{n-1}}{3^{n-1}}+\frac{x^n}{3^n}\right] \end{gathered}

Coefficient of x^n=\frac{1}{3^{n+1}}+\frac{1}{3^n}+\frac{1}{3^{n-1}}+\ldots . .(n+1) terms =\frac{1}{3^{n+1}} \frac{\left[3^{n+1}-1\right]}{3-1}=\frac{3^{n+1}-1}{2 \cdot 3^{n+1}} .

Put n=1,2,3 \ldots \ldots and find the coefficients of x, x^2, x^3 \ldots \ldots and comparing with the given option as : 

Coefficient of x^2 \text { is }=\frac{1}{3^3}+\frac{1}{3^2}+\frac{1}{3^1}=\frac{1}{3^3} \frac{\left[3^3-1\right]}{3-1}=\frac{13}{27} Which is given by option (a) \frac{3^{n+1}-1}{2 \cdot\left(3^{n+1}\right)}=\frac{3^3-1}{2 \cdot 3^3}=\frac{13}{27}

 

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