First slide

Binomial theorem for negative integral and rational Index

Question

If the expansion in powers of x of the function $1 / [(1 - ax) (1 - bx)] is a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots$ , then a_n is

Moderate

A

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

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

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

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

Solution

$\frac{1}{(1 - ax) (1 - bx)} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n} + \dots$
But $(1 - ax)^{- 1} (1 - bx)^{- 1} = (1 + ax + a^{2} x^{2} + \dots) \times (1 + bx + b^{2} x^{2} + \dots)$
$\Rightarrow Coefficient of x^{n} is$
$b^{n} + {ab}^{n - 1} + a^{2} b^{n - 2} + \dots + a^{n - 1} b + a^{n} = \frac{b^{n + 1} - a^{n + 1}}{b - a}$
$\Rightarrow a_{n} = \frac{b^{n + 1} - a^{n + 1}}{b - a}$

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