In the expansion of `(x +a)^(n)` the sum of even
terms is E and that of odd terms is O, them `O^(2) + E^(2)` is equal to

Options

A. `(x^(2)+a^(2))^(n)`
B. `(2x^(2)-a^(2))^(n)`
C. `(x^(2)-a^(2))^(2n)`
D. `(x^(2)+a^(2))^(2n)`

Answer

To solve the problem, we need to find the value of \( O^2 + E^2 \) where \( O \) is the sum of the odd terms and \( E \) is the sum of the even terms in the expansion of \( (x + a)^n \).

### Step-by-Step Solution:

1. **Understand the Expansion**:
The binomial expansion of \( (x + a)^n \) can be expressed as:
\[ (x + a)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} a^k \]
Here, the terms with even \( k \) contribute to \( E \) and the terms with odd \( k \) contribute to \( O \).

2. **Express the Sums**:
We can express the total sum of the terms as:
\[ O + E = (x + a)^n \]
and the difference of the sums as:
\[ O - E = (x - a)^n \]

3. **Set Up the Equations**:
We have two equations:
\[ O + E = (x + a)^n \quad \text{(1)} \]
\[ O - E = (x - a)^n \quad \text{(2)} \]

4. **Add and Subtract the Equations**:
By adding equations (1) and (2):
\[ (O + E) + (O - E) = (x + a)^n + (x - a)^n \]
This simplifies to:
\[ 2O = (x + a)^n + (x - a)^n \]
Thus,
\[ O = \frac{(x + a)^n + (x - a)^n}{2} \]

Now, subtracting equation (2) from equation (1):
\[ (O + E) - (O - E) = (x + a)^n - (x - a)^n \]
This simplifies to:
\[ 2E = (x + a)^n - (x - a)^n \]
Thus,
\[ E = \frac{(x + a)^n - (x - a)^n}{2} \]

5. **Calculate \( O^2 + E^2 \)**:
We can use the identity:
\[ O^2 + E^2 = \frac{1}{2} \left( (O + E)^2 + (O - E)^2 \right) \]
Substituting the values we found:
\[ O^2 + E^2 = \frac{1}{2} \left( ((x + a)^n)^2 + ((x - a)^n)^2 \right) \]

6. **Final Expression**:
Therefore, we have:
\[ O^2 + E^2 = \frac{1}{2} \left( (x + a)^{2n} + (x - a)^{2n} \right) \]

### Conclusion:
The final result for \( O^2 + E^2 \) is:
\[ O^2 + E^2 = \frac{1}{2} \left( (x + a)^{2n} + (x - a)^{2n} \right) \]