The coefficient of $x "$ in the expansion of $(1 - 9 x + 20 x^{2})^{- 1}$ is

5^{n} - 4^{n}

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5^{n + 1} - 4^{n + 1}

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5^{n - 1} - 4^{n - 1}

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None of these

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Solution

The correct option is B $5^{n + 1} - 4^{n + 1}$
We have,
$(1 - 9 x + 20 x^{2})^{- 1} = [(1 - 5 x) (1 - 4 x)]^{- 1}$

$= \frac{1}{(1 - 5 x) (1 - 4 x)} = \frac{5}{1 - 5 x} - \frac{4}{1 - 4 x}$
$5 (1 - 5 x)^{- 1} - 4 (1 - 4 x)^{- 1}$

$5 [1 + 5 x + (5 x)^{2} + . . . + (5 x)^{n} + . . . . .] - 4 [1 + 4 x + (4 x)^{2} + . . . + (4 x)^{n} + . . .]$
The coefficient of $x^{n}$ is $5^{n + 1} - 4^{n + 1}$ .

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Similar questions

Find the value of $^{n} C_{0}$ 4 + $4^{2}$ × $\frac{^{n} C_{1}}{2}$ + ......... $\frac{4^{n + 1}}{n + 1}$ × $^{n} C_{n}$

1, ω, ω^{2}, \dots, ω^{n - 1}

n^{th}

x^{n} = 1.

(5 - ω) (5 - ω^{2}) \dots (5 - ω^{n - 1})

4 C_{0} + \frac{4^{2}}{2} . C_{1} + \frac{4^{3}}{3} C_{2} + . . . . . . . . . . . . + \frac{4^{n + 1}}{n + 1} C_{n} = \frac{5^{n + 1} - 1}{n + 1}

A = {\begin{matrix} 5^{n} - 4 n - 1 : n \in N \end{matrix}}

B = {\begin{matrix} 16 (n - 1) : n \in N \end{matrix}}

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