The coefficient of {x^n} in the expansion of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We have,$(1 - 9x + 20{x^2})^{-1} = [(1 - 5x)(1 - 4x)]^{-1}$.Using partial fractions,we write:$\frac{1}{(1 - 5x)(1 - 4x)} = \frac{A}{1 - 5x} + \fra

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We have,$(1 - 9x + 20{x^2})^{-1} = [(1 - 5x)(1 - 4x)]^{-1}$.Using partial fractions,we write:$\frac{1}{(1 - 5x)(1 - 4x)} = \frac{A}{1 - 5x} + \frac{B}{1 - 4x}$.Solving for $A$ and $B$,we get $A = 5$ and $B = -4$.Thus,the expression becomes:$5(1 - 5x)^{-1} - 4(1 - 4x)^{-1}$.Expanding using the binomial series $(1 - z)^{-1} = \sum_{k=0}^{\infty} z^k$:$= 5 \sum_{k=0}^{\infty} (5x)^k - 4 \sum_{k=0}^{\infty} (4x)^k$.The coefficient of ${x^n}$ is:$5(5^n) - 4(4^n) = 5^{n+1} - 4^{n+1}$.

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

Similar Questions

If $x$ is positive,the first negative term in the expansion of $(1 + x)^{27/5}$ is

$1+\frac{4}{15}+\frac{4 \times 10}{15 \times 30}+\frac{4 \times 10 \times 16}{15 \times 30 \times 45}+\ldots \quad \infty=$

If $(a + bx)^{-2} = \frac{1}{4} - 3x + \dots$,then $(a, b) = $

The sum of the series $\frac{3}{4 \cdot 8}-\frac{3 \cdot 5}{4 \cdot 8 \cdot 12}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16}-\ldots$

If $T_4$ represents the $4^{th}$ term in the expansion of $\left(5x + \frac{7}{x}\right)^{-3/2}$ and $x \notin \left[-\sqrt{\frac{7}{5}}, \sqrt{\frac{7}{5}}\right]$,then $\left(x^7 \sqrt{5x}\right) T_4 =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module