If the expansion in powers of $x$ of the function $\frac{1}{(1-a x)(1-b x)}$ is $a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots$, then $a_n$ is

$\frac{b^{\mathrm{n}}-a^n}{b-a}$

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

If the expansion in powers of

x

of the function

\frac{1}{(1-a x)(1-b x)}

a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots

, then

a_n

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it represents the coefficient of

x^n

in the expansion of the given function using partial fractions or series expansion.

Practice All Chapter MCQs →

\left({ }^{10} C_0\right)^2-\left({ }^{10} C_1\right)^2+\ldots-\left({ }^{10} C_9\right)^2+\left({ }^{10} C_{10}\right)^2

a

d

\left( {n + 1} \right)

a{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - ... + ...

(1-x)^{201}\left(1+x+x^2\right)^{200}

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

{x^5}

\frac{{1 + {x^2}}}{{1 + x}},\left| x \right| < 1,

\left(\begin{array}{c}2 n+1 \\ 0\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 3\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 6\end{array}\right)+\ldots=170

n