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