Find the sum `5^(2)+ 6^(2) + 7^(2) + ... + 20^(2)`.

Answer

To find the sum \(5^2 + 6^2 + 7^2 + \ldots + 20^2\), we can follow these steps:

### Step 1: Identify the series
The series starts from \(5^2\) and ends at \(20^2\). The terms can be expressed as \(n^2\) where \(n\) varies from \(5\) to \(20\).

### Step 2: Determine the number of terms
To find the number of terms in the series, we can calculate:
- The first term is \(5\) and the last term is \(20\).
- The number of terms \(N\) can be calculated as:
\[ N = 20 - 5 + 1 = 16 \]

### Step 3: Use the formula for the sum of squares
The sum of squares of the first \(n\) natural numbers is given by the formula:
\[ \text{Sum} = \frac{n(n + 1)(2n + 1)}{6} \]
We will use this formula to find the sum of squares from \(1^2\) to \(20^2\) and then subtract the sum of squares from \(1^2\) to \(4^2\).

### Step 4: Calculate the sum of squares from \(1^2\) to \(20^2\)
Using the formula for \(n = 20\):
\[ \text{Sum}_{1 \text{ to } 20} = \frac{20(20 + 1)(2 \cdot 20 + 1)}{6} = \frac{20 \cdot 21 \cdot 41}{6} \]
Calculating this step-by-step:
- \(20 \cdot 21 = 420\)
- \(2 \cdot 20 + 1 = 41\)
- Now, calculate \(420 \cdot 41\):
\[ 420 \cdot 41 = 17220 \]
- Finally, divide by \(6\):
\[ \frac{17220}{6} = 2870 \]

### Step 5: Calculate the sum of squares from \(1^2\) to \(4^2\)
Using the formula for \(n = 4\):
\[ \text{Sum}_{1 \text{ to } 4} = \frac{4(4 + 1)(2 \cdot 4 + 1)}{6} = \frac{4 \cdot 5 \cdot 9}{6} \]
Calculating this step-by-step:
- \(4 \cdot 5 = 20\)
- \(2 \cdot 4 + 1 = 9\)
- Now, calculate \(20 \cdot 9\):
\[ 20 \cdot 9 = 180 \]
- Finally, divide by \(6\):
\[ \frac{180}{6} = 30 \]

### Step 6: Subtract the two sums to find the desired sum
Now, subtract the sum from \(1^2\) to \(4^2\) from the sum from \(1^2\) to \(20^2\):
\[ \text{Sum}_{5^2 \text{ to } 20^2} = \text{Sum}_{1 \text{ to } 20} - \text{Sum}_{1 \text{ to } 4} = 2870 - 30 = 2840 \]

### Final Answer
Thus, the sum \(5^2 + 6^2 + 7^2 + \ldots + 20^2 = 2840\).

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