A) \[\frac{{{a}^{n}}-{{b}^{n}}}{b-a}\]
B) \[\frac{{{a}^{n+1}}-{{b}^{n+1}}}{b-a}\]
C) \[\frac{{{b}^{n+1}}-{{a}^{n+1}}}{b-a}\]
D) \[\frac{{{b}^{n}}-{{a}^{n}}}{b-a}\]
Correct Answer: C
Solution :
Since,\[{{(1-ax)}^{-1}}{{(1-bx)}^{-1}}\] \[(1-ax+{{a}^{2}}{{x}^{2}}+.....)(1+bx+{{b}^{2}}{{x}^{2}}+....)\] Hence,\[{{a}_{n}}=\]Coefficient of\[{{x}^{n}}\]in\[{{(1-ax)}^{-1}}{{(1-bx)}^{-1}}\] \[={{a}^{0}}{{b}^{n}}+a{{b}^{n-1}}+.......{{a}^{n}}{{b}^{0}}\] \[={{a}^{0}}{{b}^{n}}\left[ \frac{{{\left( \frac{a}{b} \right)}^{n+1}}-1}{\frac{a}{b}-1} \right]\] \[=\frac{{{b}^{n}}({{a}^{n+1}}-{{b}^{n+1}})}{a-b}.\frac{b}{{{b}^{n+1}}}\] \[=\frac{{{a}^{n+1}}-{{b}^{n+1}}}{a-b}=\frac{{{b}^{n+1}}-{{a}^{n+1}}}{b-a}\]
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