TS EAMCET · Maths · Binomial Theorem
For \(|x| < \frac{1}{2}\), if the coefficient of \(x^{10}\) and the constant term in the expansion of \(\frac{2 x^3+8 x^2-2 x-2}{(1-x)(1+x)(1-2 x)}\) in powers of \(x\) are \(l\) and \(m\) respectively, then \(1 m=\)
- A \(6\left(1+2^9\right)\)
- B \(4\left(1+2^9\right)\)
- C \(6\left(1+2^{10}\right)\)
- D \(4\left(1+2^{10}\right)\)
Answer & Solution
Correct Answer
(B) \(4\left(1+2^9\right)\)
Step-by-step Solution
Detailed explanation
The given expression is \(\frac{2 x^3+8 x^2-2 x-2}{(1-x)(1+x)(1-2 x)}=\frac{2 x^3+8 x^2-2 x-2}{(2 x-1)(x-1)(x+1)}\) \(=1+\frac{A}{2 x-1}+\frac{B}{x-1}+\frac{C}{x+1}\) (By partial fraction)…
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