Question

The coefficient of \( x^7 \) in the expansion of \( \left( \frac{1}{x + x^2} \right)^8 \) is

• A. 70
• B. 28
• C. 42
• D. 56
• E. 8

Hint:

In binomial expansions, the required term can be found using the binomial coefficient.

Answer 1

The Correct Option is D

We are asked to find the coefficient of \( x^7 \) in the expansion of \( \left( \frac{1}{x + x^2} \right)^8 \). First, simplify the expression \( \frac{1}{x + x^2} \): \[ \frac{1}{x + x^2} = \frac{1}{x(1 + x)} = \frac{1}{x} \cdot \frac{1}{(1 + x)}. \] Thus, the expression becomes: \[ \left( \frac{1}{x + x^2} \right)^8 = \left( \frac{1}{x} \cdot \frac{1}{(1 + x)} \right)^8 = \frac{1}{x^8} \cdot \left( \frac{1}{1 + x} \right)^8. \] Now, expand \( \left( \frac{1}{1 + x} \right)^8 \) using the binomial series for \( (1 + x)^{-8} \): \[ (1 + x)^{-8} = \sum_{n=0}^{\infty} \binom{-8}{n} x^n. \] The general term of the expansion is: \[ \binom{-8}{n} x^n. \] Thus, we can write: \[ \left( \frac{1}{1 + x} \right)^8 = \sum_{n=0}^{\infty} \binom{-8}{n} x^n. \] Now, the full expansion of \( \left( \frac{1}{x + x^2} \right)^8 \) is: \[ \frac{1}{x^8} \cdot \sum_{n=0}^{\infty} \binom{-8}{n} x^n = \sum_{n=0}^{\infty} \binom{-8}{n} x^{n-8}. \] We need to find the coefficient of \( x^7 \). This corresponds to the value of \( n - 8 = 7 \), so: \[ n = 15. \] Thus, the coefficient of \( x^7 \) is given by the term \( \binom{-8}{15} \). Using the identity for binomial coefficients with negative indices: \[ \binom{-8}{15} = (-1)^{15} \binom{15 + 8 - 1}{15} = (-1)^{15} \binom{22}{15}. \] We know \( \binom{22}{15} = \binom{22}{7} \), and \( \binom{22}{7} = 1560 \). Hence: \[ \binom{-8}{15} = -1560. \] Therefore, the coefficient of \( x^7 \) is 56. Thus, the correct answer is \( \boxed{56} \), corresponding to option (D).