AP EAMCET · Maths · Binomial Theorem
In the expansion of \(\frac{2 x+1}{(1+x)(1-2 x)}\), the sum of the coefficients of the first 5 odd powers of \(x\) is
- A \(\frac{5}{3}+\frac{8}{9}\left(4^5-1\right)\)
- B \(\frac{5}{3}+\frac{8}{3}\left(4^5-1\right)\)
- C \(-\frac{5}{3}+\frac{8}{9}\left(4^5-1\right)\)
- D \(\frac{5}{3}+\frac{8}{12}\left(4^5+1\right)\)
Answer & Solution
Correct Answer
(A) \(\frac{5}{3}+\frac{8}{9}\left(4^5-1\right)\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Let } \frac{2 x+1}{(1+x)(1-2 x)} \\ & =\frac{A}{1+x}+\frac{B}{1-2 x}=\frac{A(1-2 x)+B(1+x)}{(1+x)(1-2 x)} \\ & \Rightarrow 2 x+1=A(1-2 x)+B(1+x)\end{aligned}\) On comparing both sides, we get…
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