Question

$1+\frac{3}{2!}+\frac{7}{3!}+\frac{15}{4!}+...\infty =?$

• A. $e(e+1)$
• B. $e(1-e)$
• C. $e(e-1)$
• D. $3e$

Answer 1

The correct option is C

$e(e-1)$

Explanation for the correct option:

Step 1. Finding $T_r$ term of the given series.

we can write each term of

$1+\frac{3}{2!}+\frac{7}{3!}+\frac{15}{4!}+...\infty$ as

$\begin{array}{rcl}{a}_1& =& 1\\ & =& \frac{2^1-1}{1!}\end{array}$,

$\begin{array}{rcl}{a}_2& =& \frac{3}{2!}\\ & =& \frac{2^2-1}{2!}\end{array}$,

$\begin{array}{rcl}{a}_3& =& \frac{7}{3!}\\ & =& \frac{2^3-1}{3!}\end{array}$ and so on

$\Rightarrow T_{\mathrm{r}}=\frac{2^{\mathrm{r}}-1}{\mathrm{r}!}$

Step 2. Find the sum of $T_r$,

$\Rightarrow \underset{r=1}{\overset{\infty }{\sum \left(T_r\right)}}=\sum _{r=1}^{\infty }\left(\frac{2^r-1}{r!}\right)$ ……..(A)

As we know,

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

$\Rightarrow e^x=1+\sum _{r=1}^{\infty }\left(\frac{x^r}{r!}\right)$

Step 3. Using the expansion.

Put $x=2$ and $r=1$ in $e^x$expansion:

$e^2=1+\sum _{r=1}^{\infty }\left(\frac{2^r}{r!}\right)$

$\Rightarrow \sum _{r=1}^{\infty }\left(\frac{2^r}{r!}\right)=e^2-1........\left(i\right)$

$e^1=1+\sum _{r=1}^{\infty }\left(\frac{1}{r!}\right)$

$\Rightarrow \sum _{r=1}^{\infty }\left(\frac{1}{r!}\right)=e^1-1$ ……….(2)

Put the values of equation (1) and (2) in equation (A):

$\Rightarrow \sum _{r=1}^{\infty }\left(T_r\right)=\left(e^2-1\right)-\left(e^1-1\right)$

$=e^2-e$

$=e(\mathrm{e}-1)$

Hence, Option $\left(C\right)$ is Correct.