MCQ
Suppose A and B are the coefficients of $30^{\text {th }}$ and $12^{\text {th }}$ terms respectively in the binomial expansion of $(1+x)^{2 n-1}$. If $2 A=5 B$, then $n$ is equal to:
- A22
- ✓21
- C20
- D19
✓
Answer
Correct option: B.
21
(B)
Sol. $\mathrm{A}={ }^{2 \mathrm{n}-1} \mathrm{C}_{29} \quad \mathrm{~B}={ }^{2 \mathrm{n}-1} \mathrm{C}_{11}$
$2{ }^{2 n-1} C_{29}=5{ }^{2 n-1} C_{11}$
$2 \frac{(2 n-1)!}{29!(2 n-30)!}=5 \frac{(2 n-1)!}{(2 n-12)!11!}$
$\frac{1}{29 \ldots 12 \cdot 5}=\frac{1}{(2 n-12)(2 n-13) \ldots(2 n-29)^{2}}$
$\frac{1}{30 \cdot 29 \ldots 12}=\frac{1}{(2 n-12)(2 n-13) \ldots(2 n-29) 12}$
$2 \mathrm{n}-12=30$
$\mathrm{n}=21$
Sol. $\mathrm{A}={ }^{2 \mathrm{n}-1} \mathrm{C}_{29} \quad \mathrm{~B}={ }^{2 \mathrm{n}-1} \mathrm{C}_{11}$
$2{ }^{2 n-1} C_{29}=5{ }^{2 n-1} C_{11}$
$2 \frac{(2 n-1)!}{29!(2 n-30)!}=5 \frac{(2 n-1)!}{(2 n-12)!11!}$
$\frac{1}{29 \ldots 12 \cdot 5}=\frac{1}{(2 n-12)(2 n-13) \ldots(2 n-29)^{2}}$
$\frac{1}{30 \cdot 29 \ldots 12}=\frac{1}{(2 n-12)(2 n-13) \ldots(2 n-29) 12}$
$2 \mathrm{n}-12=30$
$\mathrm{n}=21$
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