The sum of the series `1+(3)/(2!)+(7)/(3!)+(15)/(4!)+. . .` is

Options

A. `e(e+1_`
B. `e(1-e)`
C. `e(e-1)`
D. 3e

Answer

To find the sum of the series \( S = 1 + \frac{3}{2!} + \frac{7}{3!} + \frac{15}{4!} + \ldots \), we will first identify the general term of the series and then sum it.

### Step 1: Identify the General Term

The series can be expressed as:
\[ S = \sum_{r=1}^{\infty} \frac{t_r}{r!} \]
where \( t_r \) is the numerator of each term. Observing the numerators:
- For \( r=1 \): \( t_1 = 1 \)
- For \( r=2 \): \( t_2 = 3 \)
- For \( r=3 \): \( t_3 = 7 \)
- For \( r=4 \): \( t_4 = 15 \)

We can see a pattern in the numerators:
- \( t_1 = 1 = 2^1 - 1 \)
- \( t_2 = 3 = 2^2 - 1 \)
- \( t_3 = 7 = 2^3 - 1 \)
- \( t_4 = 15 = 2^4 - 1 \)

Thus, we can generalize:
\[ t_r = 2^r - 1 \]

### Step 2: Rewrite the Series

Now we can rewrite the series \( S \):
\[ S = \sum_{r=1}^{\infty} \frac{2^r - 1}{r!} \]
This can be split into two separate sums:
\[ S = \sum_{r=1}^{\infty} \frac{2^r}{r!} - \sum_{r=1}^{\infty} \frac{1}{r!} \]

### Step 3: Evaluate Each Sum

The first sum \( \sum_{r=1}^{\infty} \frac{2^r}{r!} \) is the Taylor series expansion for \( e^x \) evaluated at \( x = 2 \):
\[ \sum_{r=0}^{\infty} \frac{2^r}{r!} = e^2 \]
Since we start from \( r=1 \), we need to subtract the \( r=0 \) term:
\[ \sum_{r=1}^{\infty} \frac{2^r}{r!} = e^2 - 1 \]

The second sum \( \sum_{r=1}^{\infty} \frac{1}{r!} \) is also the Taylor series for \( e^x \) evaluated at \( x = 1 \):
\[ \sum_{r=0}^{\infty} \frac{1}{r!} = e \]
Again, subtracting the \( r=0 \) term:
\[ \sum_{r=1}^{\infty} \frac{1}{r!} = e - 1 \]

### Step 4: Combine the Results

Now substituting back into our expression for \( S \):
\[ S = (e^2 - 1) - (e - 1) \]
This simplifies to:
\[ S = e^2 - e \]

### Final Answer

Thus, the sum of the series is:
\[ \boxed{e^2 - e} \]