Question

Let ${(1+x)}^n=1+a_{1}x+a_1{x}^2+....+a_n{x}^n$ If $a_1$, $a_2$ and $a_3$ are in A.P., then the value of n is

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

The correct option is D $7$

$(1+x^2)=1+a_{1}x+a_2{x}^2+...+a_n{x}^n...\left(1\right)$

$\therefore$ Expanding the binomial Expression,

$(1+x{)}^n{=}^n{C}_0(1{)}^n\left(x{)}^0{+}^n{C}_1\right(1{)}^{n-1}\left(x\right){+}^n{C}_2\left(1{)}^{n-2}\right(x^2)+...(2)$

on equating (1) & (2) , we get

$\Rightarrow a_{1}x{=}^n{C}_{1}x$

$\Rightarrow a_{1}x=n.x$

$a_1=x...\left(3\right)$

and

$\Rightarrow a_2{x}^2{=}^n{C}_2{x}^2$

$\Rightarrow a_2{x}^2=\frac{n(n-1)}{2\times 1}{x}^3\Rightarrow a_2=\frac{n(n-1)}{2}...\left(4\right)$

and $a_3{x}^3{=}^n{C}_3{x}^3$

$a_3{x}^3=\frac{n(n-1)(n-2)x^3}{3\times 2\times 1}\Rightarrow a_3=\frac{n(n-1)(n-2)}{2\times 3}...\left(5\right)$

since $a_1,a_2,a_3$ are in A.P

$\therefore$ from (3),(4),(5)

$\Rightarrow 2a_2=a_1+a_3$

$\Rightarrow \frac{2\left(n\right(n-1\left)\right)}{2}=\frac{n+n(n-1)(n-2)}{6}$

$0=n^3-3n+2+12-6n$

$0=n^2-9n+14$

$(n-7)(n-2)$

$\therefore n=7,$ (option C)