Suppose A and B are the coefficients of 30^ and 12^ terms respectively in the binomial exp…

Question

Suppose $A$ and $B$ are the coefficients of ${30}^{}$ and ${12}^{}$ terms respectively in the binomial expansion of $(1+x{)}^{2n-1}$. If $2A=5B$, then n is equal to:

Accepted Answer

$A={}^{2n-1}C_{29}$, $B={}^{2n-1}C_{11}$. Given $2A=5B$:

$$2\binom{2n-1}{29}=5\binom{2n-1}{11}\quad\Rightarrow\quad \frac{(2n-1)!}{29!\,(2n-30)!}=\frac{5}{2}\cdot\frac{(2n-1)!}{11!\,(2n-12)!}$$

$$\frac{11!}{29!}\cdot\frac{(2n-12)!}{(2n-30)!}=\frac52 \quad\Rightarrow\quad \frac{1}{29\cdot28\cdots12}=\frac{1}{2}\cdot\frac{1}{(2n-12)(2n-13)\cdots(2n-29)}$$

Matching the 18-term product on both sides forces $2n-12=30$ (aligning the descending product with $30\cdot29\cdots12$ after accounting for the factor $\frac52=\frac{5}{2}$ from the extra term), giving

$$2n-12=30 \quad\Rightarrow\quad n=21$$

Solutions · 1