If P be the sum of odd terms and Q be the sum of even terms in the expansion of `(x+a)^(n)`, then `(x+a)^(2n)+(x-a)^(2n)` is
(i) `P^(2)-Q^(2)`
(ii) `P^(2)+Q^(2)`
(iii) `2(P^(2)+Q^(2))`
(iv) `4PQ`

Options

A. `P^(2)-Q^(2)`
B. `P^(2)+Q^(2)`
C. `2(P^(2)+Q^(2))`
D. `4PQ`

Answer

To solve the problem, we need to find the expression for \((x + a)^{2n} + (x - a)^{2n}\) in terms of the sums of the odd and even terms \(P\) and \(Q\) from the expansion of \((x + a)^{n}\).

### Step-by-Step Solution:

1. **Understand the Expansion**:
The binomial expansion of \((x + a)^{n}\) is given by:
\[ (x + a)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} a^{k} \]
The terms where \(k\) is odd contribute to the sum \(P\) and where \(k\) is even contribute to the sum \(Q\).

2. **Expansion of \((x + a)^{2n}\) and \((x - a)^{2n}\)**:
We can write:
\[ (x + a)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} x^{2n-k} a^{k} \]
\[ (x - a)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} x^{2n-k} (-a)^{k} \]

3. **Adding the Two Expansions**:
When we add these two expansions:
\[ (x + a)^{2n} + (x - a)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} x^{2n-k} a^{k} + \sum_{k=0}^{2n} \binom{2n}{k} x^{2n-k} (-a)^{k} \]
The odd terms (where \(k\) is odd) will cancel out, and we will be left with:
\[ = 2 \sum_{k \text{ even}} \binom{2n}{k} x^{2n-k} a^{k} \]
This sum represents \(2Q\) (the sum of even terms).

4. **Using \(P\) and \(Q\)**:
From the previous steps, we can express:
\[ (x + a)^{2n} + (x - a)^{2n} = 2Q \]

5. **Finding \(P\) and \(Q\)**:
We know:
\[ P = \frac{(x + a)^{n} + (x - a)^{n}}{2} \]
\[ Q = \frac{(x + a)^{n} - (x - a)^{n}}{2} \]

6. **Calculating \(PQ\)**:
We can find \(PQ\) as:
\[ PQ = \left(\frac{(x + a)^{n} + (x - a)^{n}}{2}\right) \left(\frac{(x + a)^{n} - (x - a)^{n}}{2}\right) = \frac{(x + a)^{2n} - (x - a)^{2n}}{4} \]

7. **Final Expression**:
Therefore, we have:
\[ (x + a)^{2n} + (x - a)^{2n} = 4PQ \]

### Conclusion:
The correct answer is:
\[ \text{(iv) } 4PQ \]