The sum of first n terms of the series `1^(2) + 2.2^(2) + 3^(2) + 2.4^(2) + 5^(2) + 2.6^(2)+…` is `(n (n + 1)^(2))/(2)` when n is even. When n is odd the sum is

Options

A. `(n (n + 1))/(2)`
B. `(n^(2) (n + 1))/(2)`
C. `(n (n + 1)^(2))/(2)`
D. `[(n(n + 1))/(2)]^(2)`

Answer

To solve the problem, we need to find the sum of the first \( n \) terms of the series given by \( 1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \ldots \) when \( n \) is odd.

### Step-by-Step Solution:

1. **Identify the Pattern**:
The series alternates between terms that are multiplied by 2 and those that are not. Specifically, the odd-indexed terms (1st, 3rd, 5th, ...) are \( k^2 \) for \( k = 1, 3, 5, \ldots \) and the even-indexed terms (2nd, 4th, 6th, ...) are \( 2 \cdot k^2 \) for \( k = 2, 4, 6, \ldots \).

2. **Express \( n \) in Terms of \( m \)**:
For odd \( n \), we can express \( n \) as \( n = 2m + 1 \), where \( m \) is a non-negative integer.

3. **Write the Sum**:
The sum \( S \) of the first \( n \) terms can be broken down into:
\[ S = 1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \ldots + (2m + 1)^2 \]
This includes \( m + 1 \) odd terms and \( m \) even terms.

4. **Calculate the Sum of Odd Terms**:
The odd terms are \( 1^2, 3^2, 5^2, \ldots, (2m + 1)^2 \).
The sum of the squares of the first \( m + 1 \) odd numbers is given by:
\[ \text{Sum of odd squares} = (m + 1)^2(2m + 1) = \frac{(m + 1)(2m + 1)(2m + 2)}{6} \]

5. **Calculate the Sum of Even Terms**:
The even terms are \( 2 \cdot 2^2, 2 \cdot 4^2, \ldots, 2 \cdot (2m)^2 \).
The sum of the squares of the first \( m \) even numbers is:
\[ \text{Sum of even squares} = 2 \cdot (2^2 + 4^2 + 6^2 + \ldots + (2m)^2) = 2 \cdot 4 \cdot \frac{m(m + 1)(2m + 1)}{6} = \frac{4m(m + 1)(2m + 1)}{6} \]

6. **Combine the Sums**:
Now, combine the sums of the odd and even terms:
\[ S = \frac{(m + 1)(2m + 1)(2m + 2)}{6} + \frac{4m(m + 1)(2m + 1)}{6} \]

7. **Simplify the Expression**:
Factor out common terms:
\[ S = \frac{(m + 1)(2m + 1)}{6} \left( (2m + 2) + 4m \right) = \frac{(m + 1)(2m + 1)(6m + 2)}{6} \]

8. **Final Expression**:
Simplifying gives:
\[ S = \frac{(m + 1)(2m + 1)(3m + 1)}{3} \]

### Final Result:
Thus, the sum of the first \( n \) terms when \( n \) is odd is:
\[ S = \frac{(n)(n + 1)(n + 2)}{6} \]