The coefficient of `x^(4)` in the expansion of `(1+5x+9x^(2)+13x^(3)+17x^(4)+…..)(1+x^(2))^(11)` is equal to

Options

A. `.^(11)C_(2)+4.^(11)C_(1)+3`
B. `.^(11)C_(2)+3.^(11)C_(1)+4`
C. `3.^(11)C_(2)+4.^(11)C_(1)+3`
D. 171

Answer

To find the coefficient of \(x^4\) in the expansion of the expression \((1 + 5x + 9x^2 + 13x^3 + 17x^4 + \ldots)(1 + x^2)^{11}\), we will break down the problem step by step.

### Step 1: Identify the first part of the expression
The first part of the expression is a polynomial:
\[ 1 + 5x + 9x^2 + 13x^3 + 17x^4 + \ldots \]
We can observe that the coefficients \(5, 9, 13, 17\) are in an arithmetic sequence with a common difference of \(4\). The general term for this sequence can be expressed as:
\[ a_k = 4k + 1 \]
where \(k\) starts from \(0\).

### Step 2: Write the general term
The general term of the polynomial can be written as:
\[ a_k x^k = (4k + 1)x^k \]
Thus, the polynomial can be expressed as:
\[ \sum_{k=0}^{4} (4k + 1)x^k \]

### Step 3: Expand the second part of the expression
The second part of the expression is:
\[ (1 + x^2)^{11} \]
Using the binomial theorem, we can expand this as:
\[ (1 + x^2)^{11} = \sum_{j=0}^{11} \binom{11}{j} (x^2)^j = \sum_{j=0}^{11} \binom{11}{j} x^{2j} \]

### Step 4: Find the coefficient of \(x^4\)
To find the coefficient of \(x^4\) in the product of the two expansions, we need to consider the combinations of terms from each expansion that will result in \(x^4\).

1. **From the first polynomial**:
- The constant term \(1\) (coefficient from \(k=0\)) contributes to \(x^4\) when multiplied by the \(x^4\) term from the binomial expansion.
- The term \(5x\) (coefficient from \(k=1\)) contributes to \(x^3\) when multiplied by the \(x^2\) term from the binomial expansion.
- The term \(9x^2\) (coefficient from \(k=2\)) contributes to \(x^2\) when multiplied by the \(x^2\) term from the binomial expansion.
- The term \(13x^3\) (coefficient from \(k=3\)) contributes to \(x^1\) when multiplied by the constant term from the binomial expansion.
- The term \(17x^4\) (coefficient from \(k=4\)) contributes to \(x^0\) when multiplied by the constant term from the binomial expansion.

2. **Calculating contributions**:
- From \(1\) and \(x^4\): Coefficient is \(\binom{11}{2} = 55\)
- From \(5x\) and \(x^2\): Coefficient is \(5 \cdot \binom{11}{1} = 5 \cdot 11 = 55\)
- From \(9x^2\) and \(x^2\): Coefficient is \(9 \cdot \binom{11}{2} = 9 \cdot 55 = 495\)
- From \(13x^3\) and \(1\): Coefficient is \(13 \cdot \binom{11}{0} = 13\)
- From \(17x^4\) and \(1\): Coefficient is \(17\)

### Step 5: Sum the contributions
Now, we add all the contributions:
\[ 55 + 55 + 495 + 13 + 17 = 635 \]

### Final Answer
The coefficient of \(x^4\) in the expansion is:
\[ \boxed{635} \]