Question

If the ratio of the coefficient of third and fourth term in the expansion of $(x-\frac{1}{2x}{)}^n$ is $1:2$, then the value of $n$ will be

• A. 18
• B. 16
• C. 12
• D. -10

Answer 1

Answer: (D)

In binomial expansion $(x+y{)}^n$,

General term $=T_{r+1}{=}^n{C}_r.x^{n-r}.y^r$

Now, in $(x-\frac{1}{2x}{)}^n$

$T_3{=}^n{C}_2\left(x{)}^{n-2}\right(-\frac{1}{2x}{)}^2$

$\Rightarrow T_3{=}^n{C}_2\left(x{)}^{n-2}\right(\frac{1}{2^2{x}^2})$

$\Rightarrow T_3{=}^n{C}_2\left(x{)}^{n-2-2}\right(\frac{1}{2^2})$

$\Rightarrow T_3{=}^n{C}_2\left(x{)}^{n-4}\right(\frac{1}{4})$---(1)

$T_4{=}^n{C}_3\left(x{)}^{n-3}\right(-\frac{1}{2x}{)}^3$

$\Rightarrow T_3{=}^n{C}_3\left(x{)}^{n-2}\right(-\frac{1}{2^3{x}^3})$

$\Rightarrow T_3={-}^n{C}_3\left(x{)}^{n-2-3}\right(\frac{1}{8})$

$\Rightarrow T_3={-}^n{C}_3\left(x{)}^{n-5}\right(\frac{1}{8})$---(2)

It is given that, the ratio of the coefficient of third and fourth term is $1:2$

$\Rightarrow \frac{{}^n{C}_2.\frac{1}{4}}{{-}^n{C}_3.\frac{1}{8}}=\frac{1}{2}$

$\Rightarrow \frac{{}^n{C}_2.\frac{1}{4}}{{}^n{C}_3.\frac{1}{8}}=-\frac{1}{2}$

$\Rightarrow \frac{\frac{n(n-1)}{2.1}}{\frac{n(n-1)(n-2)}{\mathrm{3.2.1}}}.2=-\frac{1}{2}$

$\Rightarrow \frac{n(n-1)}{2.1}\times \frac{\mathrm{3.2.1}}{n(n-1)(n-2)}.2=-\frac{1}{2}$

$\Rightarrow \frac{6}{n-2}=-\frac{1}{2}$

$\Rightarrow 12=-n+2$

$\Rightarrow n=2-12$

$\Rightarrow n=-10$

Hence, Option C is correct.