Question

The coefficient of $x^n$ in $(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...-\frac{(-1){}^{n}x{}^n}{n!}){}^{2}sequalto$

• A. $\frac{\left(n\right){}^n}{n!}$
• B. $\frac{(-1{)}^n(2){}^n}{n!}$
• C. $\frac{1}{(n!){}^2}$
• D. $-\frac{1}{(n!){}^2}$

Answer 1

The correct option is B $\frac{(-1{)}^n(2){}^n}{n!}$

Hint: Try using the fact that $e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots -\frac{(-1{)}^n{x}^n}{n!}$.

Step 1: Simplication of the given series:

The series expansion of $e^{-x}$ is given by $e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots -\frac{(-1{)}^n{x}^n}{n!}$.

Note that the given expression is the square of $1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots -\frac{(-1{)}^n{x}^n}{n!}$.

The square of $1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots -\frac{(-1{)}^n{x}^n}{n!}$ would correspond to $(e^{-x}{)}^2$, that is, $e^{-2x}$.

Substitute $2x$ for $x$ in $e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots -\frac{(-1{)}^n{x}^n}{n!}$ to obtain the series expansion of $e^{-2x}$.

$\begin{array}{cc}{e}^{-2x}& =1-2x+\frac{(2x{)}^2}{2!}-\frac{(2x{)}^3}{3!}+\cdots -\frac{(-1{)}^n(2x{)}^n}{n!}\\ e^{-2x}& =1-2x+\frac{4x^2}{2!}-\frac{8x^3}{3!}+\cdots -\frac{(-1{)}^n(2{)}^n{x}^n}{n!}\end{array}$

Step 2 : Calculation of the nth term of the series.

The coefficient of the $n$th term in $e^{-2x}=1-2x+\frac{4x^2}{2!}-\frac{8x^3}{3!}+\cdots -\frac{(-1{)}^n{2}^n{x}^n}{n!}$ is $\frac{(-1{)}^n(2{)}^n}{n!}$.

Final step: The $n$th term of the the given series is $\frac{(-1{)}^n(2{)}^n}{n!}$.