Question

The coefficient $x^n$ in the expression of ${\left(1+x\right)}^{2n}$ and ${\left(1+x\right)}^{2n-1}$ are in the ratio.

• A. $1:2$
• B. $1:3$
• C. $3:1$
• D. $2:1$

Answer 1

The correct option is A $1:2$

We know that,

General term of $(a+b{)}^n$ is

$T_r+1{=}^n{C}_r{a}^{n-r},b^r$

For $(1+x{)}^{2n}$

General term of $(1+x{)}^{2n}$

Put $a=1,b=x,n=2n$

$T_r+1{=}^{2n}{C}_r(1{)}^{2n-r}.(x{)}^r$

$T{Y}_r+1{=}^{2n}{C}_r.(x{)}^r$......(i)

For coedfficient of $x^n$

Put $r=n$ in $e{q}^n$(i)

$T_r+1{=}^{2n}{C}_n{x}^n$

Coefficient of $x^n{=}^{2n}{C}_n$

$\therefore 2^n{C}_n=\frac{2n!}{n!(2n-n)!}$

$=\frac{2n!}{n!\left(n\right)!}$

For $(1+x{)}^{2n-1}$

General term of $(1+x{)}^{2n-1}$

Put $a=1,b=x,n=2n-1$

$T_r+1{=}^{2r-1}{C}^r.(1{)}^{2n-1-r}.x^r$

$T_r+1{=}^{2n-1}{C}_r{x}^r$.....(ii)

For coefficient of $x^n$

Put $r=n$ in equation (ii)

$T_r+1{=}^{2n-1}{C}_n{x}^n$

Coefficient of $x^n$

$=2^{n-1}{C}_n\times 2$

$=2\times \frac{(2n-1)!}{n!(2n-1-n)!}$

$=2\times \frac{(2n-1)!}{n!(n-1)!}$

Multiply and divide by $n$,

we get,

$=\frac{2n(2n-1)!}{n!n(n-1)!}$

$=\frac{\left(2n\right)!}{n!n!}$

Hence, ration is $1:2$