Question

The coefficient of $x^n$ in expansion of ${\left(1+2x+3x^2+....\right)}^{\frac{1}{2}}$ is

• A. $-1$
• B. $0$
• C. $2$
• D. $1$

Answer 1

The correct option is D

$1$

Explanation for the correct option:

Step 1: Solving using the standard expansion

Given expression is ${\left(1+2x+3x^2+....\right)}^{\frac{1}{2}}$

The standard expansion of ${\left(1-x\right)}^{-n}$ is given by

${\left(1-x\right)}^{-n}=1+nx+\frac{n\left(n+1\right)}{2!}{x}^2+\frac{n\left(n+1\right)\left(n+2\right)}{3!}{x}^3+...$

$\Rightarrow {\left(1-x\right)}^{-2}=\left(1+2x+3x^2+....\right)$

Taking square root on both sides,

$\Rightarrow {\left(1+2x+3x^2+....\right)}^{\frac{1}{2}}={\left[{\left(1-x\right)}^{-2}\right]}^{\frac{1}{2}}$

$\Rightarrow {\left(1+2x+3x^2+....\right)}^{\frac{1}{2}}={\left(1-x\right)}^{-1}$

Step 2: Determining the coefficient of $x^n$

Now, using Identity ${\left(1-x\right)}^{-1}$

We know, ${\left(1-x\right)}^{-1}=\left(1+x+x^2+x^3+......+x^n+...\infty \right)$

$\therefore {\left(1+2x+3x^2+....\right)}^{\frac{1}{2}}=\left(1+x+x^2+....+x^n+...\infty \right)$

On comparing, we see that the coefficient of $x^n$ is $1$.

Hence, option (D) is the correct answer.