What is the coefficient of x^4 in the expansi | vedclass

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

Question

(B) First,we express the given fraction using partial fractions: $\frac{3x}{(x - 2)(x + 1)} = \frac{A}{x - 2} + \frac{B}{x + 1}$ $3x = A(x + 1) + B(x

Vedclass · Accepted Answer

$\frac{15}{16}$

Vedclass · Answer

(B) First,we express the given fraction using partial fractions: $\frac{3x}{(x - 2)(x + 1)} = \frac{A}{x - 2} + \frac{B}{x + 1}$ $3x = A(x + 1) + B(x - 2)$ For $x = 2$,$6 = 3A \implies A = 2$. For $x = -1$,$-3 = -3B \implies B = 1$. So,$\frac{3x}{(x - 2)(x + 1)} = \frac{2}{x - 2} + \frac{1}{x + 1} = \frac{2}{-2(1 - x/2)} + \frac{1}{1 + x} = -(1 - x/2)^{-1} + (1 + x)^{-1}$. Using the binomial expansion $(1 - z)^{-1} = 1 + z + z^2 + z^3 + z^4 + \dots$ and $(1 + z)^{-1} = 1 - z + z^2 - z^3 + z^4 - \dots$: The expression becomes $-(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \frac{x^4}{16} + \dots) + (1 - x + x^2 - x^3 + x^4 - \dots)$. The coefficient of $x^4$ is $-\frac{1}{16} + 1 = \frac{15}{16}$.

What is the coefficient of $x^4$ in the expansion of $\frac{3x}{(x - 2)(x + 1)}$?

Similar Questions

If $\frac{1}{(3-5 x)(2+3 x)}=\frac{A}{3-5 x}+\frac{B}{2+3 x}$,then $A+B$ is equal to

The partial fraction of $\frac{x^2 - 5}{x^2 - 3x + 2}$ is:

If $\frac{3x + 4}{{(x + 1)}^2(x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{{(x + 1)}^2}$,then $A = $

If $\frac{\sin^2 x + 1}{2\sin^2 x - 5\sin x + 3} = \frac{A}{2\sin x - 3} + \frac{B}{\sin x - 1} + C$,then:

$\begin{aligned} & \text{If } \frac{x^4}{(x-a)(x-b)(x-c)}=P(x)+\frac{A}{x-a}+\frac{B}{x-b} \\ & +\frac{C}{x-c} \text{, then } P(0)+A(a-b)(a-c)= \end{aligned}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module