Let x be a real number and -2

Question

(A) Using partial fractions,we write: $\frac{x+1}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}$.Solving for $A$ and $B$,we get $A = \frac{2}{5}$ and $B

Vedclass · Accepted Answer

$-\frac{55}{1296}$

Vedclass · Answer

(A) Using partial fractions,we write: $\frac{x+1}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}$.Solving for $A$ and $B$,we get $A = \frac{2}{5}$ and $B = \frac{3}{5}$.Thus,$\frac{x+1}{(x+3)(x-2)} = \frac{2}{5(x+3)} + \frac{3}{5(x-2)} = \frac{2}{15(1 + x/3)} - \frac{3}{10(1 - x/2)}$.Using the binomial expansion $(1+u)^{-1} = 1 - u + u^2 - u^3 + \dots$ and $(1-u)^{-1} = 1 + u + u^2 + u^3 + \dots$,we have:$= \frac{2}{15} \left(1 - \frac{x}{3} + \frac{x^2}{9} - \frac{x^3}{27} + \dots ight) - \frac{3}{10} \left(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \dots ight)$.The coefficient of $x^3$ is $\frac{2}{15} 	imes (-\frac{1}{27}) - \frac{3}{10} 	imes \frac{1}{8} = -\frac{2}{405} - \frac{3}{80} = -\frac{32 + 243}{6480} = -\frac{275}{6480} = -\frac{55}{1296}$.

Let $x$ be a real number and $-2 < x < 2$. When $\frac{x+1}{(x+3)(x-2)}$ is expanded in powers of $x$,then the coefficient of $x^3$ is

Similar Questions

$\frac{1}{x(x+1)(x+2) \ldots(x+n)} = \frac{A_0}{x} + \frac{A_1}{x+1} + \ldots + \frac{A_n}{x+n}$. For $0 \leq r \leq n$,$A_r$ is equal to:

$\frac{4x^2+5}{(x-2)^4} = \frac{A}{(x-2)} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)^3} + \frac{D}{(x-2)^4}$,then $\sqrt{\frac{A}{C} + \frac{B}{C} + \frac{D}{C}} = $

If $\frac{2x^3+x^2-5}{x^4-25}=\frac{Ax+B}{x^2-5}+\frac{Cx+1}{x^2+5}$, then $(A, B, C)$ equals to

If $\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}=\frac{A x+B}{x^2+1}+\frac{C x+D}{\left(x^2+1\right)^2}+\frac{E x+F}{\left(x^2+1\right)^3}$,then the value of $A+B+C+D+E+F=$

If $\frac{1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$ and $\frac{x}{(x-1)(x-2)(x-3)}=\frac{P}{x-1}+\frac{Q}{x-2}+\frac{R}{x-3}$, then $A+2B+3C=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module