Question

Let $n$ be a positive integer. If the coefficients of $2^{nd},3^{rd}$ and $4^{th}$ terms in the expansion of $(1+x{)}^n$ are in G.P., then the value of $n$ is:

• A. $4$
• B. $3$
• C. $5$
• D. None of the above.

Answer 1

The correct option is D None of the above.

The coefficent of the genral term can be written as $T_{r+1}$=${}^n{C}_r$

$T_{1+1}$ $=$ ${}^n{C}_1$ $=n$

$T_{2+1}$ $=$ ${}^n{C}_2$ $=$ $\frac{n(n-1)}{2}$

$T_{3+1}$ $=$ ${}^n{C}_3$ $=$ $\frac{n(n-1)(n-2)}{6}$

As $T_2$, $T_3$ and $T_4$ are in G.P.

$\Rightarrow$${T_3}^2$ $=$ $T_2.$$T_4$

$\Rightarrow$$\frac{n^2{(n-1)}^2}{4}$ $=$ $\frac{n^2(n-1)(n-2)}{6}$

$\Rightarrow$$n(n-1)(\frac{n^2-n}{4}-\frac{n^2-2n}{6})=0$

$\Rightarrow$$n(n-1)(n^2+n)=0$

$\Rightarrow$$n^2(n-1)(n+1)=0$

So, $n$ can be $0,1$ or $-1$.

$n$ can't be $0$ or $-1$ as $n$ is a positive integer.

$n$ can't be $1$ as there are $2^{nd},3^{rd}$ and $4^{th}$ terms.