Kerala KEAM 5 June 2024 Solved Paper

We are asked to find the coefficient of $x^3$ in the expansion of $\frac{1}{(1 + 2x)^{-10}}$.We can write the expression as: $$\frac{1}{(1 + 2x)^{-10}} = (1 + 2x)^{10}$$Now, apply the binomial expansion to $(1 + 2x)^{10}$. The binomial expansion for $(1 + 2x)^n$ is given by: $$(1 + 2x)^{10} = \sum_{k=0}^{10} \binom{10}{k} (2x)^k$$We need to find the coefficient of $x^3$. This corresponds to the term where $k = 3$ in the expansion: $$\binom{10}{3} (2x)^3$$ $$= \binom{10}{3} \cdot 2^3 \cdot x^3$$Now, calculate $\binom{10}{3}$ and $2^3$: $$\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$$ $$2^3 = 8$$Thus, the coefficient of $x^3$ is: $$120 \times 8 = 960$$Thus, the correct answer is option (B), 960. Quick Tip: When dealing with binomial expansions of the form $(1 + ax)^n$, use the binomial theorem and identify the term corresponding to the power of $x$ you are interested in. Remember that $\binom{n}{k} a^k x^k$ represents the general term.