Question

$\frac{2}{1!}+\frac{6}{2!}+\frac{12}{3!}+\frac{20}{4!}+......\infty$

• A. $\frac{3e}{2}$
• B. $e$
• C. $2e$
• D. $3e$

Answer 1

The correct option is D

$3e$

Explanation for correct option:

Step-1: Simplify the given sum

Let the given series,

$\mathrm{S}=\frac{2}{1!}+\frac{6}{2!}+\frac{12}{3!}+\frac{20}{4!}+......\infty$

$$\begin{gathered} \Rightarrow \mathrm{S}=\frac{1\times 2}{1!}+\frac{2\times 3}{2!}+\frac{3\times 4}{3!}+\frac{4\times 5}{4!}+......\infty \\ \Rightarrow \mathrm{S}=2+\frac{3}{1!}+\frac{4}{2!}+\frac{5}{3!}+......\infty .....\left[1\right] \end{gathered}$$

Step 2: Use $e^x$ expansion formula

We know that,

$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+........\infty$

Multiply by $x^2$ both the side

$\Rightarrow x^2{e}^x=x^2+\frac{x^3}{1!}+\frac{x^4}{2!}+........\infty$

Differentiating both sides with respect to $x$

$2x{e}^x+x^2{e}^x=2x+\frac{3x^2}{1!}+\frac{4x^3}{2!}+......\infty$

Step 3: Find the required result

Substitute, $x=1$.

$$\begin{gathered} 2\times 1\times e^1+{\left(1\right)}^2{e}^1=2\times 1+\frac{3{\left(1\right)}^2}{1!}+\frac{4{\left(1\right)}^3}{2!}+\frac{5{\left(1\right)}^4}{3!}......\infty \\ \Rightarrow 3e=2+\frac{3}{1!}+\frac{4}{2!}+\frac{5}{3!}......\infty \end{gathered}$$

$\Rightarrow 3e=\frac{2}{1!}+\frac{6}{2!}+\frac{12}{3!}+\frac{20}{4!}+......\infty$

Comparing with equation [1]

$S=3e$

Hence, correct answer is option $\left(D\right)$