Let $r > 1$ and $n > 2$ be integers. Suppose $L$ and $M$ are the coefficients of $(3 r)^{t h}$ and $(r + 2)^{t h}$
terms respectively in the binomial expansion of $2 n (1 + x)^{23} - 1$ . If $(r + 2) L = 3 r (M)$ then $n$ is

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$2 r + 1$

$2 r - 1$

$2 r$

$2 r + 2$

answer is C.

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Detailed Solution

Clearly ,

$L = 2 n \times^{2 n - 1} C_{3 r - 1}$ and $M = 2 n^{2 n - 1} C_{r + 1}$

$\begin{array}{ll} ∴ (r + 2) L = 3 r (M) \\ \Rightarrow (r + 2) 2 n \times 2 n - 1 C_{3 r - 1} = 3 r \times 2 n + r - 1 C_{r + 1} \\ \Rightarrow \frac{(2 n - 1)! (r + 2)}{(3 r - 1)! (2 n - 3 r)!} = \frac{3 r \times (2 n - 1)!}{(2 n - r - 2)! (r + 1)!} \end{array}$

$\begin{array}{l} = \frac{1}{3 r! (2 n - 3 r)!} = \frac{1}{(r + 2)! (2 n - r - 2)!} \\ = \frac{(2 n)!}{3 r! (2 n - 3 r)!} = \frac{(2 n)!}{(r + 2)! (2 n - r - 2)!} \\ \Rightarrow^{2 n} C_{3 r} =^{2 n} C_{r + 2} \\ \Rightarrow 3 r + r + 2 = 2 n \Rightarrow n = 2 r + 1 \end{array}$