If the coefficient of `x^8` in the expansion of `(1+ (x^2)/( 2!) + (x^4)/( 4!) + (x^6)/( 6!) + (x^8)/( 8!))^(2)` is `(1)/(M)`, then a divisor of `M` is

Options

A. 2
B. 3
C. 5
D. 7

Answer

To find the coefficient of \( x^8 \) in the expansion of

\[ \left(1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \frac{x^8}{8!}\right)^2, \]

we can start by recognizing that the series inside the parentheses is the Taylor series expansion for \( e^x \) but only including even powers of \( x \).

### Step 1: Recognizing the Series
The series can be rewritten as:

\[ \frac{e^x + e^{-x}}{2} = \cosh(x). \]

Thus, we have:

\[ \left( \cosh(x) \right)^2. \]

### Step 2: Using the Identity
Using the identity \( \cosh^2(x) = \frac{1 + \cos(2x)}{2} \), we can express our function as:

\[ \cosh^2(x) = \frac{1 + \cos(2x)}{2}. \]

### Step 3: Finding the Coefficient of \( x^8 \)
Next, we need to find the coefficient of \( x^8 \) in the expansion of \( \cosh^2(x) \). The series expansion for \( \cosh(x) \) is:

\[ \cosh(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \frac{x^8}{8!} + \ldots \]

When we square this series, we need to consider the terms that contribute to \( x^8 \):

1. \( \left( \frac{x^8}{8!} \right) \) from \( \cosh(x) \) squared.
2. \( 2 \cdot \left( \frac{x^4}{4!} \cdot \frac{x^4}{4!} \right) \) which gives \( \frac{2x^8}{(4!)^2} \).
3. \( 2 \cdot \left( \frac{x^2}{2!} \cdot \frac{x^6}{6!} \right) \) which gives \( \frac{2x^8}{2! \cdot 6!} \).

### Step 4: Calculating Each Contribution
Now, we can calculate each contribution:

1. From \( \frac{x^8}{8!} \): Coefficient is \( \frac{1}{8!} \).
2. From \( 2 \cdot \frac{x^4}{4!} \cdot \frac{x^4}{4!} \): Coefficient is \( 2 \cdot \frac{1}{(4!)^2} = \frac{2}{16} = \frac{1}{8} \).
3. From \( 2 \cdot \frac{x^2}{2!} \cdot \frac{x^6}{6!} \): Coefficient is \( 2 \cdot \frac{1}{2! \cdot 6!} = \frac{2}{2 \cdot 720} = \frac{1}{720} \).

### Step 5: Combining Contributions
Now, we combine these contributions to find the total coefficient of \( x^8 \):

\[ \text{Total Coefficient} = \frac{1}{8!} + \frac{1}{8} + \frac{1}{720}. \]

Calculating \( 8! = 40320 \), we get:

\[ \frac{1}{8!} = \frac{1}{40320}, \quad \frac{1}{8} = \frac{5040}{40320}, \quad \frac{1}{720} = \frac{56}{40320}. \]

Adding these gives:

\[ \frac{1 + 5040 + 56}{40320} = \frac{5097}{40320}. \]

### Step 6: Finding \( M \)
Given that the coefficient of \( x^8 \) is \( \frac{1}{M} \), we have:

\[ M = \frac{40320}{5097}. \]

### Step 7: Finding a Divisor of \( M \)
To find a divisor of \( M \), we can factor \( 40320 \) and \( 5097 \).

The prime factorization of \( 40320 \) is \( 2^7 \cdot 3^2 \cdot 5 \cdot 7 \).

The prime factorization of \( 5097 \) can be calculated or checked for divisibility.

### Conclusion
The divisors of \( M \) include \( 1, 3, 5, 7, 9, 15, 21, 35, 63, 105, 126, 315, 504, 720, 1008, 1680, 2520, 5040, 40320 \).

Thus, a divisor of \( M \) could be \( 3, 5, \) or \( 7 \).