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`

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**:
The left-hand side is \( (4x^2 + 1)^n \). We can expand this 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. **Identifying the Right-Hand Side**:
The right-hand side is given as:
\[ \sum_{r=0}^{n} a_r (1 + x^2)^{n-r} x^{2r} \]
Here, \( (1 + x^2)^{n-r} \) can also be expanded using the Binomial Theorem:
\[ (1 + x^2)^{n-r} = \sum_{j=0}^{n-r} \binom{n-r}{j} (x^2)^j \]

3. **Combining the Terms**:
Substituting this expansion back into the right-hand side gives:
\[ \sum_{r=0}^{n} a_r \left( \sum_{j=0}^{n-r} \binom{n-r}{j} x^{2j} \right) x^{2r} \]
This can be rearranged to:
\[ \sum_{r=0}^{n} a_r \sum_{j=0}^{n-r} \binom{n-r}{j} x^{2(j+r)} \]

4. **Matching Coefficients**:
For the two sides to be equal, the coefficients of \( x^{2k} \) must match for each \( k \). On the left-hand side, the coefficient of \( x^{2k} \) is \( \binom{n}{k} 4^k \). On the right-hand side, we need to consider all combinations of \( r \) and \( j \) such that \( j + r = k \).

5. **Finding \( \sum_{r=0}^{n} a_r \)**:
To find \( \sum_{r=0}^{n} a_r \), we can evaluate the original expression at \( x = 1 \):
\[ (4(1^2) + 1)^n = (4 + 1)^n = 5^n \]
The right-hand side becomes:
\[ \sum_{r=0}^{n} a_r (1 + 1^2)^{n-r} 1^{2r} = \sum_{r=0}^{n} a_r (2)^{n-r} = \sum_{r=0}^{n} a_r 2^{n-r} \]

6. **Setting Up the Equation**:
Thus, we have:
\[ 5^n = \sum_{r=0}^{n} a_r 2^{n-r} \]

7. **Finding \( \sum_{r=0}^{n} a_r \)**:
If we let \( x = 1 \) in the equation \( \sum_{r=0}^{n} a_r 2^{n-r} \), we can express it as:
\[ \sum_{r=0}^{n} a_r = \sum_{r=0}^{n} a_r 2^0 = \sum_{r=0}^{n} a_r \]
Therefore, we can deduce that:
\[ \sum_{r=0}^{n} a_r = 5^n / 2^n = \left( \frac{5}{2} \right)^n \]

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

\[ \sum_{r=0}^{n} a_r = \left( \frac{5}{2} \right)^n \]