Find the sum of the series `(2^2-1)(6^2-1)+(4^2-1)(8^2-1)+...+(100^2-1)(104^2-1)`

Answer

To find the sum of the series \((2^2-1)(6^2-1)+(4^2-1)(8^2-1)+...+(100^2-1)(104^2-1)\), we will follow these steps:

### Step 1: Rewrite the terms
The series can be rewritten as:
\[ (2^2 - 1)(6^2 - 1) + (4^2 - 1)(8^2 - 1) + \ldots + (100^2 - 1)(104^2 - 1) \]
This can be expressed as:
\[ (2^2 - 1)(6^2 - 1) = (4 - 1)(36 - 1) = 3 \cdot 35 \]
\[ (4^2 - 1)(8^2 - 1) = (16 - 1)(64 - 1) = 15 \cdot 63 \]
Continuing this pattern, we can see that each term can be expressed as:
\[ (2n^2 - 1)(6n^2 - 1) \text{ for } n = 1, 2, 3, \ldots, 50 \]

### Step 2: General term formulation
The general term can be expressed as:
\[ (2n^2 - 1)(6n^2 - 1) = (2n^2 - 1)(6n^2 - 1) = 12n^4 - 8n^2 - 6n^2 + 1 = 12n^4 - 14n^2 + 1 \]

### Step 3: Sum the series
Now we need to sum this expression from \(n = 1\) to \(n = 50\):
\[ S = \sum_{n=1}^{50} (12n^4 - 14n^2 + 1) \]
This can be separated into three sums:
\[ S = 12\sum_{n=1}^{50} n^4 - 14\sum_{n=1}^{50} n^2 + \sum_{n=1}^{50} 1 \]

### Step 4: Calculate the individual sums
Using the formulas for the sums:
- \(\sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6}\)
- \(\sum_{n=1}^{k} n^4 = \frac{k(k+1)(2k+1)(3k^2 + 3k - 1)}{30}\)
- \(\sum_{n=1}^{k} 1 = k\)

For \(k = 50\):
1. Calculate \(\sum_{n=1}^{50} n^2\):
\[ \sum_{n=1}^{50} n^2 = \frac{50 \cdot 51 \cdot 101}{6} = 42925 \]

2. Calculate \(\sum_{n=1}^{50} n^4\):
\[ \sum_{n=1}^{50} n^4 = \frac{50 \cdot 51 \cdot 101 \cdot (3 \cdot 50^2 + 3 \cdot 50 - 1)}{30} \]
First calculate \(3 \cdot 50^2 + 3 \cdot 50 - 1 = 7500 + 150 - 1 = 7649\):
\[ \sum_{n=1}^{50} n^4 = \frac{50 \cdot 51 \cdot 101 \cdot 7649}{30} = 42925 \cdot 7649 / 30 = 108900 \]

3. Calculate \(\sum_{n=1}^{50} 1 = 50\).

### Step 5: Substitute back into the sum
Now substituting back into the expression for \(S\):
\[ S = 12 \cdot 108900 - 14 \cdot 42925 + 50 \]
Calculating each term:
\[ S = 1306800 - 600950 + 50 = 706800 \]

### Final Answer
Thus, the sum of the series is:
\[ \boxed{706800} \]