The sum ` sum_(n=1)^(10) ( n(2n-1)(2n+1))/( 5) ` is equal to _.

Explanation

To find the sum

\[ S = \sum_{n=1}^{10} \frac{n(2n-1)(2n+1)}{5}, \]

we can simplify the expression inside the summation first.

### Step 1: Simplifying the expression
The term \( n(2n-1)(2n+1) \) can be expanded. We know that \( (2n-1)(2n+1) = 4n^2 - 1 \) (this is a difference of squares). Therefore, we can rewrite the term as:

\[ n(2n-1)(2n+1) = n(4n^2 - 1) = 4n^3 - n. \]

Thus, we can express the sum \( S \) as:

\[ S = \frac{1}{5} \sum_{n=1}^{10} (4n^3 - n). \]

### Step 2: Separating the summation
Now we can separate the summation:

\[ S = \frac{1}{5} \left( 4 \sum_{n=1}^{10} n^3 - \sum_{n=1}^{10} n \right). \]

### Step 3: Using summation formulas
We will use the formulas for the summation of the first \( n \) natural numbers and the summation of the cubes of the first \( n \) natural numbers:

1. The sum of the first \( n \) natural numbers is given by:

\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2}. \]

2. The sum of the cubes of the first \( n \) natural numbers is given by:

\[ \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2. \]

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 \( \sum_{n=1}^{10} n^3 \):

\[ \sum_{n=1}^{10} n^3 = \left( \frac{10(10+1)}{2} \right)^2 = \left( \frac{10 \times 11}{2} \right)^2 = 55^2 = 3025. \]

### Step 4: Plugging values back into the equation
Now substituting these values back into our expression for \( S \):

\[ S = \frac{1}{5} \left( 4 \cdot 3025 - 55 \right). \]

Calculating inside the parentheses:

\[ 4 \cdot 3025 = 12100, \]
so,

\[ S = \frac{1}{5} (12100 - 55) = \frac{1}{5} \cdot 12045. \]

### Step 5: Final calculation
Now, divide by 5:

\[ S = 2409. \]

Thus, the final answer is:

\[ \boxed{2409}. \]