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

(D) $\frac{b^{n+1}-a^{n+1}}{b-a}$

(A) $\frac{b^{n}-a^{n}}{b-a}$

(B) $\frac{a^{n}-b^{n}}{b-a}$

(C) $\frac{a^{n+1}-b^{n+1}}{b-a}$

JEE Main · 2006 · 30 Apr 2006 · Offline · Binomial Theorem

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} + \dots

, then

a_{n}

Answer: (D)

Expand each factor as a geometric series:

\frac{1}{1 - a x} = 1 + a x + a^{2} x^{2} + \dots

and

\frac{1}{1 - b x} = 1 + b x + b^{2} x^{2} + \dots

. Multiplying, the coefficient of

x^{n}

collects every product

a^{i} b^{j}

with

i + j = n

, namely

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

. This is a geometric series of

n + 1

terms, and summing it gives

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

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option