The coefficient of `x^(7)` in the expansion of `(1+1/(1!)x+1/(2!)x^(2)+1/(3!)x^(3)+1/(4!)x^(4)+1/(5!)x^(5))^(2)` is:

Options

A. `2/(7!)(""^(8)C_(3))`
B. `(2^(10))/(5!)`
C. `4/15`
D. `7/(20!)`

Answer

To find the coefficient of \( x^7 \) in the expansion of

\[ \left( 1 + \frac{1}{1!} x + \frac{1}{2!} x^2 + \frac{1}{3!} x^3 + \frac{1}{4!} x^4 + \frac{1}{5!} x^5 \right)^2, \]

we can follow these steps:

### Step 1: Identify the series
The expression inside the parentheses is a finite series that can be rewritten as:

\[ \sum_{k=0}^{5} \frac{1}{k!} x^k. \]

### Step 2: Expand the series
We need to find the coefficient of \( x^7 \) in the square of this series. This means we need to consider the products of terms from the series that will yield \( x^7 \).

### Step 3: Determine combinations of terms
To obtain \( x^7 \), we can consider the following combinations of terms from the series:

1. \( x^5 \) from the first series and \( x^2 \) from the second series.
2. \( x^4 \) from the first series and \( x^3 \) from the second series.
3. \( x^3 \) from the first series and \( x^4 \) from the second series.
4. \( x^2 \) from the first series and \( x^5 \) from the second series.

However, since the series only goes up to \( x^5 \), the combinations of \( x^3 \) and \( x^4 \) will be the same as \( x^4 \) and \( x^3 \), and we can ignore duplicates.

### Step 4: Calculate the coefficients
Now, we calculate the coefficients for the valid combinations:

1. **For \( x^5 \) and \( x^2 \)**:
- Coefficient of \( x^5 \) is \( \frac{1}{5!} \).
- Coefficient of \( x^2 \) is \( \frac{1}{2!} \).
- Contribution:
\[ 2 \cdot \frac{1}{5!} \cdot \frac{1}{2!} = 2 \cdot \frac{1}{120} \cdot \frac{1}{2} = \frac{1}{120}. \]

2. **For \( x^4 \) and \( x^3 \)**:
- Coefficient of \( x^4 \) is \( \frac{1}{4!} \).
- Coefficient of \( x^3 \) is \( \frac{1}{3!} \).
- Contribution:
\[ 2 \cdot \frac{1}{4!} \cdot \frac{1}{3!} = 2 \cdot \frac{1}{24} \cdot \frac{1}{6} = \frac{1}{72}. \]

### Step 5: Sum the contributions
Now, we sum the contributions from both combinations:

\[ \frac{1}{120} + \frac{1}{72}. \]

To add these fractions, we find a common denominator. The least common multiple of 120 and 72 is 360.

Converting \( \frac{1}{120} \) to have a denominator of 360:

\[ \frac{1}{120} = \frac{3}{360}. \]

Converting \( \frac{1}{72} \) to have a denominator of 360:

\[ \frac{1}{72} = \frac{5}{360}. \]

Now, adding these:

\[ \frac{3}{360} + \frac{5}{360} = \frac{8}{360} = \frac{2}{90} = \frac{1}{45}. \]

### Conclusion
Thus, the coefficient of \( x^7 \) in the expansion is

\[ \frac{1}{45}. \]