Question

The coefficient of $x^n$ in the series $1+\frac{\left(a+bx\right)}{1!}+\frac{{\left(a+bx\right)}^2}{2!}+\frac{{\left(a+bx\right)}^3}{3!}+.....$ is

• A. $\frac{{\left(ab\right)}^n}{n!}$
• B. $\frac{e^b{a}^n}{n!}$
• C. $\frac{e^a{b}^n}{n!}$
• D. $\frac{e^{a+b}{\left(ab\right)}^n}{n!}$

Answer 1

The correct option is C

$\frac{e^a{b}^n}{n!}$

Explanation for the correct answer:

Finding the coefficient of $x^n$:

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

We know that, $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+......$

Here replace $x$ with $ax+b$ we get,

$e^{\left(a+bx\right)}=1+\frac{\left(a+bx\right)}{1!}+\frac{{\left(a+bx\right)}^2}{2!}+\frac{{\left(a+bx\right)}^3}{3!}+......$

$\Rightarrow e^{\left(a+bx\right)}=e^a\left(e^{bx}\right)$

$\Rightarrow e^{\left(a+bx\right)}=e^a\left(1+bx+\frac{{\left(bx\right)}^2}{2!}+....\right)$

So, the coefficient of $x^n$in the expansion of $e^{\left(a+bx\right)}=1+\frac{\left(a+bx\right)}{1!}+\frac{{\left(a+bx\right)}^2}{2!}+\frac{{\left(a+bx\right)}^3}{3!}+......$is $\frac{e^a{b}^n}{n!}$

Hence, option (C) is the correct answer.