ExamBro
ExamBro
JEE Mains · Maths · STD 11 - 8. sequence and series

The sum \(1+\frac{1+3}{2!}+\frac{1+3+5}{3!}+\frac{1+3+5+7}{4!}+\ldots\) upto \(\infty\) terms, is equal to

  1. A \(6 e\)
  2. B \(4 e\)
  3. C \(3 e\)
  4. D \(2 e\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(2 e\)

Step-by-step Solution

Detailed explanation

\begin{aligned} & \mathrm{S}=1+\frac{1+3}{2!}+\frac{1+3+5}{3!}+\ldots \\ & =\sum_{\mathrm{r}=1}^{\infty} \frac{\mathrm{r}^2}{\mathrm{r}!} \\ & =\sum_{\mathrm{r}=1}^{\infty} \frac{(\mathrm{r}-1+1)}{(\mathrm{r}-1)!}=\sum_{\mathrm{r}=2}^{\infty}…

From JEE Mains
Explore more questions on app