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).