AP EAMCET · Maths · Binomial Theorem
Let \(x\) be a real number and \(-2 < x < 2\). When \(\frac{x \quad 1}{(x+3)(x-2)}\) is expanded in powers of \(x\), then the coefficient of \(x^3\) is
- A \(-\frac{55}{1296}\)
- B \(-\frac{97}{216}\)
- C \(-\frac{13}{216}\)
- D \(-\frac{119}{1800}\)
Answer & Solution
Correct Answer
(A) \(-\frac{55}{1296}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { } \frac{x+1}{(x+3)(x-2)}=\frac{3}{5} \frac{1}{(x-2)}+\frac{2}{5} \frac{1}{(x+3)} \\ & =-\frac{3}{5 \times 2}\left(1-\frac{x}{2}\right)^{-1}+\frac{2}{5 \times 3}\left(1+\frac{x}{3}\right)^{-1} \\ & \Rightarrow…
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