The coefficient of x20 in the expansion of `(1+x^(2))^(40)*(x^(2)+2+1/x^(2))^(-5)` is :

Options

A. `""^(30)C_(10)`
B. `""^(30)C_(25)`
C. 1
D. None of these

Answer

To find the coefficient of \( x^{20} \) in the expansion of \( (1 + x^2)^{40} \cdot (x^2 + 2 + \frac{1}{x^2})^{-5} \), we will follow these steps:

### Step 1: Simplify the expression
We start with the expression:
\[ (1 + x^2)^{40} \cdot (x^2 + 2 + \frac{1}{x^2})^{-5} \]
The second part can be rewritten as:
\[ (x^2 + 2 + \frac{1}{x^2}) = \left( x + \frac{1}{x} \right)^2 + 2 \]
Thus, we can express:
\[ (x^2 + 2 + \frac{1}{x^2})^{-5} = \left( \left( x + \frac{1}{x} \right)^2 + 2 \right)^{-5} \]

### Step 2: Expand \( (1 + x^2)^{40} \)
Using the Binomial Theorem, we can expand \( (1 + x^2)^{40} \):
\[ (1 + x^2)^{40} = \sum_{k=0}^{40} \binom{40}{k} x^{2k} \]

### Step 3: Expand \( (x^2 + 2 + \frac{1}{x^2})^{-5} \)
We can rewrite \( (x^2 + 2 + \frac{1}{x^2})^{-5} \) as:
\[ \left( \frac{1}{x^2} (1 + 2x^{-2} + 1) \right)^{-5} = x^{10} (1 + 2x^{-2} + 1)^{-5} \]
Now, we can expand \( (1 + 2 + 1/x^2)^{-5} \) using the generalized binomial expansion:
\[ (1 + 2 + 1/x^2)^{-5} = \sum_{m=0}^{\infty} \binom{-5}{m} (2 + 1/x^2)^m \]

### Step 4: Find the coefficient of \( x^{20} \)
Now, we need to find the coefficient of \( x^{20} \) from the product of the two expansions:
1. From \( (1 + x^2)^{40} \), we have terms of the form \( x^{2k} \).
2. From \( (x^2 + 2 + \frac{1}{x^2})^{-5} \), we have terms of the form \( x^{10} \cdot \text{(other terms)} \).

We need to find combinations of \( k \) and \( m \) such that:
\[ 2k + 10 + \text{(power from the second expansion)} = 20 \]

### Step 5: Calculate the coefficients
The coefficient of \( x^{20} \) can be calculated as follows:
- From \( (1 + x^2)^{40} \), we need \( 2k + 10 = 20 \) which gives \( k = 5 \).
- The coefficient from \( (1 + x^2)^{40} \) for \( k = 5 \) is \( \binom{40}{5} \).
- The remaining part must yield \( x^0 \) from the second expansion.

Thus, the coefficient of \( x^{20} \) is:
\[ \text{Coefficient} = \binom{40}{5} \]

### Final Step: Conclusion
The coefficient of \( x^{20} \) in the expansion of \( (1 + x^2)^{40} \cdot (x^2 + 2 + \frac{1}{x^2})^{-5} \) is:
\[ \binom{40}{5} \]