If |x|

Question

(B) Given,$\frac{3x}{(x-2)(x+1)}$ can be written as partial fractions: $\frac{3x}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1} \quad (i)$ $3x = A(x+1)

Vedclass · Accepted Answer

$\frac{-33}{32}$

Vedclass · Answer

(B) Given,$\frac{3x}{(x-2)(x+1)}$ can be written as partial fractions: $\frac{3x}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1} \quad (i)$ $3x = A(x+1) + B(x-2)$ At $x = 2$,$3(2) = A(2+1)$ $\Rightarrow 6 = 3A$ $\Rightarrow A = 2$. At $x = -1$,$3(-1) = B(-1-2)$ $\Rightarrow -3 = -3B$ $\Rightarrow B = 1$. Substituting $A$ and $B$ in Eq. $(i)$,we get: $\frac{3x}{(x-2)(x+1)} = \frac{2}{x-2} + \frac{1}{x+1} = \frac{2}{-2(1 - \frac{x}{2})} + \frac{1}{1+x} = -(1 - \frac{x}{2})^{-1} + (1+x)^{-1}$ Using the binomial expansion $(1-y)^{-1} = 1 + y + y^2 + y^3 + y^4 + y^5 + \dots$ and $(1+y)^{-1} = 1 - y + y^2 - y^3 + y^4 - y^5 + \dots$: $= -[1 + \frac{x}{2} + (\frac{x}{2})^2 + (\frac{x}{2})^3 + (\frac{x}{2})^4 + (\frac{x}{2})^5 + \dots] + [1 - x + x^2 - x^3 + x^4 - x^5 + \dots]$ The coefficient of $x^5$ is $-(\frac{1}{2})^5 - 1 = -\frac{1}{32} - 1 = -\frac{33}{32}$.

If $|x| < 1$,then the coefficient of $x^5$ in the expansion of $\frac{3x}{(x-2)(x+1)}$ is

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