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If the expansion in the power of x of the function   \frac{1}{(1-a x)(1-b x)}   is a_{0}+a_{1} x+a_{2} x^{2}+\dots then an is ?

Option: 1

{a_{n}=\frac{a^{n}-b^{n}}{a-b}}


Option: 2

{a_{n}=\frac{a^{n+1}-b^{n+1}}{a-b}}


Option: 3

{a_{n}=\frac{a^{n-1}-b^{n-1}}{a-b}}


Option: 4

1


Answers (1)

best_answer

Some Important Binomial Expansion

\\1.\;\;(1+x)^{-1}=1-x+x^{2}-x^{3}+\cdots\\2.\;\;(1-x)^{-1}=1+x+x^{2}+x^{3}+\cdots\\\3.\;\;(1+x)^{-2}=1-2 x+3 x^{2}-4 x^{3}+\cdots\\4.\;\;(1-x)^{-2}=1+2 x+3 x^{2}+4 x^{3}+\cdots

 

Now

(1-a x)^{-1}(1-b x)^{-1}=\left(1+a x+a^{2} x^{2}+\ldots .\right)\left(1+b x+b^{2} x^{2}+\ldots\right)

In the equation a_{0}+a_{1} x+a_{2} x^{2}+\dots ,  an is the coefficient of xn.

\begin{array}{l}{\text { Coefficient of } x^{n}=a^{n}+a^{n-1} b+\ldots+ a b^{n-1}+b^{n}} \\ \end{array}

a^{n}+a^{n-1} b+\ldots a b^{n-1}+b^{n}  is a GP with common ratio \frac{b}{a} and (n+1) terms

Sum of the series is 

\\ {\frac{a^n\left [ \left (\frac{b}{a} \right )^{n+1}-1 \right ]}{\frac{b}{a}-1}}\\\\ \frac{a^n\frac{\left [ b ^{n+1}-a^{n+1} \right ]}{a^{n+1}}{}}{\frac{b-a}{a}}\\

{a_{n}=\frac{a^{n+1}-b^{n+1}}{a-b}}

Option B is correct.

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shivangi.bhatnagar

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