Question

In the expansion of \(\frac{2x+1}{(1+x)(1-2x)}\), the sum of the coefficients of the first 5 odd powers of \(x\) is:

• A. \( \frac{5}{3} + \frac{8}{9} (45 - 1) \)
• B. \( \frac{5}{3} + \frac{8}{3} (45 - 1) \)
• C. \( \frac{-5}{3} + \frac{8}{9} (45 - 1) \)
• D. \( \frac{5}{3} + \frac{8}{12} (45 + 1) \)

Hint:

For binomial expansions, identify the terms corresponding to the required powers of \( x \) and use the binomial coefficients to find their sum.

Answer 1

The Correct Option is A

Step 1: Simplifying the expression. We are given the function: \[ \frac{2x + 1}{(1+x)(1-2x)}. \] Expanding the denominator: \[ (1+x)(1-2x) = 1 - x - 2x^2. \] So the expression becomes: \[ \frac{2x + 1}{1 - x - 2x^2}. \] Step 2: Expanding the function. We expand the denominator using the binomial series and collect the coefficients of the odd powers of \( x \). Step 3: Final result. After collecting the terms, we find that the sum of the coefficients of the first 5 odd powers of \( x \) is: \[ \frac{5}{3} + \frac{8}{9} (45 - 1). \]