The sum, `underset(n-1)overset(10)(Sigma)(n(2n+1)(2n-1))/5` is equal to.........

Explanation

To solve the summation \( \sum_{n=1}^{10} \frac{n(2n+1)(2n-1)}{5} \), we can follow these steps:

### Step 1: Rewrite the Summation
The expression inside the summation can be simplified. We know that:
\[ (2n+1)(2n-1) = 4n^2 - 1 \]
Thus, we can rewrite the summation as:
\[ \sum_{n=1}^{10} \frac{n(4n^2 - 1)}{5} \]

### Step 2: Factor Out the Constant
Since \( \frac{1}{5} \) is a constant, we can factor it out of the summation:
\[ \frac{1}{5} \sum_{n=1}^{10} n(4n^2 - 1) \]

### Step 3: Distribute the Summation
Now, we can distribute \( n \) inside the summation:
\[ \frac{1}{5} \left( \sum_{n=1}^{10} 4n^3 - \sum_{n=1}^{10} n \right) \]

### Step 4: Use Summation Formulas
We will use the formulas for the summation of cubes and the summation of the first \( n \) natural numbers:
- The formula for the sum of the first \( n \) natural numbers is:
\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \]
- The formula for the sum of the cubes of the first \( n \) natural numbers is:
\[ \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 \]

### Step 5: Calculate Each Summation
For \( n = 10 \):
- Calculate \( \sum_{n=1}^{10} n \):
\[ \sum_{n=1}^{10} n = \frac{10(10+1)}{2} = \frac{10 \times 11}{2} = 55 \]

- Calculate \( \sum_{n=1}^{10} n^3 \):
\[ \sum_{n=1}^{10} n^3 = \left( \frac{10(10+1)}{2} \right)^2 = 55^2 = 3025 \]

### Step 6: Substitute Back into the Expression
Now substitute these values back into the expression:
\[ \frac{1}{5} \left( 4 \times 3025 - 55 \right) \]

### Step 7: Simplify the Expression
Calculate:
\[ 4 \times 3025 = 12100 \]
Thus,
\[ 12100 - 55 = 12045 \]

### Step 8: Final Calculation
Now, divide by 5:
\[ \frac{12045}{5} = 2409 \]

### Final Answer
The value of the summation is:
\[ \boxed{2409} \]