If the expansion in power of $x$ of the function $\frac{1}{{\left( {1 - ax} \right)\left( {1 - bx} \right)}}\;{\rm{is}}\,\;{a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ...,\;{\rm{then\;}}{a_n}\;{\rm{is}}$

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

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

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

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Hard

1 of 1

Application +4 | -1

If the expansion in power of

x

of the function

\frac{1}{{\left( {1 - ax} \right)\left( {1 - bx} \right)}}\;\mathrm{is}\,\;{a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ...,\;\mathrm{then\;}{a_n}\;\mathrm{is}

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 by expressing the given function as a product of two series and finding the coefficient of

x^n

, we arrive at this expression.