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

Options

A. `(2)/(15)`
B. `(4)/(15)`
C. `(1)/(30)`
D. `(2)/(45)`

Answer

To find the coefficient of \( x^5 \) in the expansion of

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

we can follow these steps:

### Step 1: Identify the series
The expression inside the parentheses is a finite series. We can rewrite it as:

\[ S = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!}. \]

### Step 2: Expand the square
We need to find \( S^2 \):

\[ S^2 = \left(1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!}\right) \cdot \left(1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!}\right). \]

### Step 3: Find the coefficient of \( x^5 \)
To find the coefficient of \( x^5 \) in \( S^2 \), we need to consider all the combinations of terms from the two series that multiply to give \( x^5 \).

1. **From \( 1 \) and \( \frac{x^5}{5!} \)**:
- Coefficient: \( \frac{1}{5!} = \frac{1}{120} \)

2. **From \( \frac{x}{1!} \) and \( \frac{x^4}{4!} \)**:
- Coefficient: \( \frac{1}{1!} \cdot \frac{1}{4!} = \frac{1}{1 \cdot 24} = \frac{1}{24} \)

3. **From \( \frac{x^2}{2!} \) and \( \frac{x^3}{3!} \)**:
- Coefficient: \( \frac{1}{2!} \cdot \frac{1}{3!} = \frac{1}{2 \cdot 6} = \frac{1}{12} \)

4. **From \( \frac{x^3}{3!} \) and \( \frac{x^2}{2!} \)** (same as above):
- Coefficient: \( \frac{1}{12} \)

5. **From \( \frac{x^4}{4!} \) and \( \frac{x}{1!} \)** (same as above):
- Coefficient: \( \frac{1}{24} \)

6. **From \( \frac{x^5}{5!} \) and \( 1 \)** (same as above):
- Coefficient: \( \frac{1}{120} \)

### Step 4: Sum the coefficients
Now we sum all the coefficients:

\[ \text{Total coefficient of } x^5 = \frac{1}{120} + \frac{1}{24} + \frac{1}{12} + \frac{1}{12} + \frac{1}{24} + \frac{1}{120}. \]

### Step 5: Convert to a common denominator
The common denominator for \( 120, 24, \) and \( 12 \) is \( 120 \):

\[ \frac{1}{120} + \frac{5}{120} + \frac{10}{120} + \frac{10}{120} + \frac{5}{120} + \frac{1}{120} = \frac{1 + 5 + 10 + 10 + 5 + 1}{120} = \frac{32}{120}. \]

### Step 6: Simplify the fraction
Now we simplify \( \frac{32}{120} \):

\[ \frac{32}{120} = \frac{8}{30} = \frac{4}{15}. \]

### Final Answer
Thus, the coefficient of \( x^5 \) in the expansion is

\[ \frac{4}{15}. \]

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