`sum_(n=1)^7(n(n+1)(2n+1))/4` is equal to

Explanation

To evaluate the summation \( \sum_{n=1}^{7} \frac{n(n+1)(2n+1)}{4} \), we can follow these steps:

### Step 1: Simplify the Expression
First, we can simplify the expression inside the summation:

\[ \frac{n(n+1)(2n+1)}{4} \]

This can be rewritten as:

\[ \frac{1}{4} n(n+1)(2n+1) \]

### Step 2: Distribute and Expand
Now, we can expand \( n(n+1)(2n+1) \):

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

So, we have:

\[ \frac{1}{4} (2n^3 + 3n^2 + n) \]

### Step 3: Separate the Summation
Now we can separate the summation:

\[ \sum_{n=1}^{7} \frac{1}{4} (2n^3 + 3n^2 + n) = \frac{1}{4} \left( 2 \sum_{n=1}^{7} n^3 + 3 \sum_{n=1}^{7} n^2 + \sum_{n=1}^{7} n \right) \]

### Step 4: Use Summation Formulas
We will use the following formulas for the summations:
- \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \)
- \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \)
- \( \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 \)

For \( n = 7 \):
1. \( \sum_{n=1}^{7} n = \frac{7 \cdot 8}{2} = 28 \)
2. \( \sum_{n=1}^{7} n^2 = \frac{7 \cdot 8 \cdot 15}{6} = 140 \)
3. \( \sum_{n=1}^{7} n^3 = \left( \frac{7 \cdot 8}{2} \right)^2 = 784 \)

### Step 5: Substitute Back into the Summation
Now we substitute these values back into our expression:

\[ \frac{1}{4} \left( 2 \cdot 784 + 3 \cdot 140 + 28 \right) \]

Calculating each term:

- \( 2 \cdot 784 = 1568 \)
- \( 3 \cdot 140 = 420 \)
- \( 28 = 28 \)

Adding these together:

\[ 1568 + 420 + 28 = 2016 \]

### Step 6: Final Calculation
Now, we divide by 4:

\[ \frac{2016}{4} = 504 \]

Thus, the final result is:

\[ \sum_{n=1}^{7} \frac{n(n+1)(2n+1)}{4} = 504 \]