If `(4x^(2) + 1)^(n) = sum_(r=0)^(n)a_(r)(1+x^(2))^(n-r)x^(2r)`, then the value of `sum_(r=0)^(n)a_(r)` is

Options

A. `3^(n)`
B. `4^(n)`
C. `5^(n)`
D. `6^(n)`

Explanation

To solve the problem, we need to find the value of \( \sum_{r=0}^{n} a_r \) given the equation:

\[ (4x^2 + 1)^n = \sum_{r=0}^{n} a_r (1 + x^2)^{n-r} x^{2r} \]

### Step-by-Step Solution:

1. **Understanding the Given Equation**:
We start with the equation \( (4x^2 + 1)^n \). This can be expressed using the binomial theorem:
\[ (4x^2 + 1)^n = \sum_{k=0}^{n} \binom{n}{k} (4x^2)^k (1)^{n-k} = \sum_{k=0}^{n} \binom{n}{k} 4^k x^{2k} \]

2. **Rewriting the Right-Hand Side**:
The right-hand side of the equation is given as:
\[ \sum_{r=0}^{n} a_r (1 + x^2)^{n-r} x^{2r} \]
We can rewrite \( (1 + x^2)^{n-r} \) using the binomial theorem as well:
\[ (1 + x^2)^{n-r} = \sum_{j=0}^{n-r} \binom{n-r}{j} x^{2j} \]

3. **Combining Both Sides**:
We need to equate both sides. The left-hand side can be expressed as:
\[ \sum_{k=0}^{n} \binom{n}{k} 4^k x^{2k} \]
And the right-hand side can be expressed as:
\[ \sum_{r=0}^{n} a_r \sum_{j=0}^{n-r} \binom{n-r}{j} x^{2(j+r)} \]

4. **Finding Coefficients**:
To find \( a_r \), we need to compare coefficients of \( x^{2m} \) on both sides. By matching the coefficients, we find:
\[ a_r = \binom{n}{r} 4^r \]

5. **Calculating \( \sum_{r=0}^{n} a_r \)**:
Now we need to find \( \sum_{r=0}^{n} a_r \):
\[ \sum_{r=0}^{n} a_r = \sum_{r=0}^{n} \binom{n}{r} 4^r \]
This is the binomial expansion of \( (1 + 4)^n \):
\[ \sum_{r=0}^{n} a_r = (1 + 4)^n = 5^n \]

### Conclusion:
Thus, the value of \( \sum_{r=0}^{n} a_r \) is:

\[ \boxed{5^n} \]