A) \[\frac{{{3}^{n+1}}-1}{{{2.3}^{n+1}}}\]
B) \[\frac{{{3}^{n+1}}-1}{{{3}^{n+1}}}\]
C) \[\left( \frac{{{3}^{n+1}}-1}{{{3}^{n+1}}} \right)\]
D) None of these
Correct Answer: A
Solution :
We have \[\frac{1}{(1-x)(3-x)}=\frac{1}{2}\left( \frac{1}{1-x}-\frac{1}{3-x} \right)\] \[=\frac{1}{2}[{{(1-x)}^{-1}}-{{(3-x)}^{-1}}]\]\[=\frac{1}{2}\left[ {{(1-x)}^{-1}}-\frac{1}{3}{{\left( 1-\frac{x}{3} \right)}^{-1}} \right]\] \[=\frac{1}{2}\left[ (1+x+{{x}^{2}}+{{x}^{3}}+...)-\frac{1}{3}\left( 1+\frac{x}{3}+{{\left( \frac{x}{3} \right)}^{2}}+{{\left( \frac{x}{3} \right)}^{3}}+... \right) \right]\] \ Coefficient of \[{{x}^{n}}\]is \[\frac{1}{2}\left[ 1-\frac{1}{3}.\frac{1}{{{3}^{n}}} \right]=\frac{1}{2}\frac{[{{3}^{n+1}}-1]}{{{3}^{n+1}}}\]
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