First slide

Binomial theorem for negative integral and rational Index

Question

If the expansion in powers of x of the function $\frac{1}{(1 - ax) (1 - bx)} is a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots$ then $α_{n}$ is

Moderate

A

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

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

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

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

Solution

$∵ (1 - ax)^{- 1} (1 - bx)^{- 1}$
$= (1 + ax + a^{2} x^{2} + \dots) (1 + bx + b^{2} x^{2} + \dots)$
$∴ a_{n} = coefficient of x^{n} in (1 - ax)^{- 1} (1 - bx)^{- 1}$
$= a^{0} b^{n} + {ab}^{n - 1} + \dots + a^{n} b^{0}$
$\begin{array}{l} = a^{0} b^{n} [1 + \frac{a}{b} + {(\frac{a}{b})}^{2} + \dots] \\ = a^{0} b^{n} [\frac{{(\frac{a}{b})}^{n + 1} - 1}{\frac{a}{b} - 1}] \\ = \frac{b^{n} (a^{n + 1} - b^{n + 1})}{a - b} \cdot \frac{b}{b^{n + 1}} \\ = \frac{a^{n + 1} - b^{n + 1}}{a - b} \end{array}$

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