QUESTION

# The expansion of ${{\text{e}}^{\text{x}}}$ is which among the following? ${\text{A}}{\text{. }}\sum\limits_{{\text{r = 0}}}^\infty {\dfrac{{{{\text{x}}^{\text{r}}}}}{{\text{r}}}} \\ {\text{B}}{\text{. }}\sum\limits_{{\text{r = 0}}}^\infty {\dfrac{{{{\text{x}}^{\text{r}}}}}{{{\text{r!}}}}} \\ {\text{C}}{\text{. }}\sum\limits_{{\text{r = 0}}}^\infty {\dfrac{{{{\text{x}}^{{\text{r + 1}}}}}}{{{\text{r + 1}}}}} \\ {\text{D}}{\text{. }}\sum\limits_{{\text{r = 0}}}^\infty {\dfrac{{{{\text{x}}^{{\text{r + 1}}}}}}{{\left( {{\text{r + 1}}} \right)!}}} \\$

Hint: To determine the expansion of ${{\text{e}}^{\text{x}}}$ , we observe the Taylor series expansion for the given term and then evaluate which among the given options is a valid general term for the expansion. The general term should hold true for all the individual terms of the expansion.

Given Data, ${{\text{e}}^{\text{x}}}$
The Taylor series expansion of the term ${{\text{e}}^{\text{x}}}$ is given as
${{\text{e}}^{\text{x}}}$ = $1 + {\text{x + }}\dfrac{{{{\text{x}}^2}}}{{2!}} + \dfrac{{{{\text{x}}^3}}}{{3!}} + \dfrac{{{{\text{x}}^4}}}{{4!}} + ........\infty$
The term $\sum\limits_{{\text{r = 0}}}^\infty {\dfrac{{{{\text{x}}^{\text{r}}}}}{{{\text{r!}}}}}$ holds true for every term of the expansion of ${{\text{e}}^{\text{x}}}$ .
For the first term of ${{\text{e}}^{\text{x}}}$ , i.e. r = 0
$\sum\limits_{{\text{r = 0}}}^\infty {\dfrac{{{{\text{x}}^{\text{r}}}}}{{{\text{r!}}}}}$ = $\sum\limits_{{\text{r = 0}}}^0 {\dfrac{{{{\text{x}}^0}}}{{{\text{0!}}}} = 1}$ --- (0! = 1 and any number to the power 0 = 1)
Hence the expansion of ${{\text{e}}^{\text{x}}}$ is $\sum\limits_{{\text{r = 0}}}^\infty {\dfrac{{{{\text{x}}^{\text{r}}}}}{{{\text{r!}}}}}$ .
Note – In order to solve this type of questions the key is to understand the concept of ‘e’ also known as the Napier’s constant used for the natural logarithm and it has a value of 2.71828. The Taylor series of a polynomial is a polynomial itself, ${{\text{e}}^{\text{x}}}$ expansion holds because the derivative of ${{\text{e}}^{\text{x}}}$ is ${{\text{e}}^{\text{x}}}$ itself and ${{\text{e}}^0}$ equals to 1.