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
- A \(6 e\)
- B \(4 e\)
- C \(3 e\)
- D \(2 e\)
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}…
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