The sum of the first n terms of the series `1^2+ 2.2^2+3^2 +2.4^2+....` is `(n(n+1)^2)/2` when n is even. Then the sum if n is odd , is

Explanation

To find the sum of the series \(1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + \ldots\) when \(n\) is odd, we start by analyzing the series and the given information.

### Step 1: Understanding the Series
The series can be expressed as:
- For even \(n\): \(S = 1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + \ldots + 2n^2\)
- The sum of the first \(n\) terms when \(n\) is even is given by:
\[ S = \frac{n(n+1)^2}{2} \]

### Step 2: Expressing the Series for Odd \(n\)
When \(n\) is odd, the last term will not be \(2n^2\) but rather \(n^2\). Therefore, we can express the sum \(S_0\) for odd \(n\) as:
\[ S_0 = 1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + \ldots + 2(n-1)^2 + n^2 \]
Here, the last even term is \(2(n-1)^2\) and the last odd term is \(n^2\).

### Step 3: Separating the Last Term
We can separate the last term \(n^2\) from the sum:
\[ S_0 = \left(1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + \ldots + 2(n-1)^2\right) + n^2 \]

### Step 4: Using the Even \(n\) Formula
The first part of the sum can be evaluated using the formula for even \(n\). Since the last even term is \(2(n-1)^2\), we replace \(n\) with \(n-1\):
\[ S_{even} = \frac{(n-1)((n-1)+1)^2}{2} = \frac{(n-1)(n)(n)}{2} \]

### Step 5: Combining the Results
Now, we can combine this with the last term \(n^2\):
\[ S_0 = \frac{(n-1)n^2}{2} + n^2 \]

### Step 6: Simplifying the Expression
To simplify, we need a common denominator:
\[ S_0 = \frac{(n-1)n^2 + 2n^2}{2} = \frac{(n-1)n^2 + 2n^2}{2} = \frac{(n^2 - n^2 + 2n^2)}{2} = \frac{(n^2 + n^2)}{2} = \frac{n(n+1)^2}{2} \]

### Final Result
Thus, the sum of the series when \(n\) is odd is:
\[ S_0 = \frac{n(n+1)^2}{2} \]