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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 \(1 m=\)

  1. A \(6\left(1+2^9\right)\)
  2. B \(4\left(1+2^9\right)\)
  3. C \(6\left(1+2^{10}\right)\)
  4. D \(4\left(1+2^{10}\right)\)
Verified Solution

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)…