Suppose $A$ and $B$ are the coefficients of $30^{\text{th}}$ and $12^{\text{th}}$ terms respectively in the binomial expansion of $(1+x)^{2n-1}$. If $2A=5B$, then $n$ is equal to:
Step-by-Step Solution
Key Concept: $A=\binom{2n-1}{29},\ B=\binom{2n-1}{11}.$ Cross-multiplying $2A=5B$ and writing the ratio as a product of $18$ telescoping factors, the only sensible solution comes from making the consecutive factor sequences offset.
$\dfrac{A}{B}=\dfrac{\binom{2n-1}{29}}{\binom{2n-1}{11}}=\dfrac{11!\,(2n-12)!}{29!\,(2n-30)!}=\dfrac{(2n-12)(2n-13)\cdots(2n-29)}{29\cdot 28\cdots 12}.$
Each is a product of $18$ terms. Setting $\dfrac{A}{B}=\dfrac{5}{2}$ and trying $2n-12=30$ (so the numerator is $30,29,\dots,13$ and denominator $29,28,\dots,12$) gives ratio $=\dfrac{30}{12}=\dfrac{5}{2}.$
So $2n-12=30\Rightarrow n=21.$
Correct Answer: 3