Find the sum of the series : `5^(2)+6^(2)+7^(2)+…+20^(2).`

Explanation

To find the sum of the series \(5^2 + 6^2 + 7^2 + \ldots + 20^2\), we can use the formula for the sum of the squares of the first \(n\) natural numbers:

\[ \text{Sum of squares} = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 1: Calculate the sum of squares from \(1^2\) to \(20^2\)

Here, \(n = 20\).

\[ \text{Sum} = \frac{20(20 + 1)(2 \times 20 + 1)}{6} \]

Calculating each part:

- \(20 + 1 = 21\)
- \(2 \times 20 + 1 = 41\)

Now substituting these values into the formula:

\[ \text{Sum} = \frac{20 \times 21 \times 41}{6} \]

### Step 2: Simplify the expression

Calculating \(20 \times 21\):

\[ 20 \times 21 = 420 \]

Now substituting back into the sum:

\[ \text{Sum} = \frac{420 \times 41}{6} \]

Calculating \(420 \div 6\):

\[ 420 \div 6 = 70 \]

Now we have:

\[ \text{Sum} = 70 \times 41 \]

### Step 3: Final multiplication

Calculating \(70 \times 41\):

\[ 70 \times 41 = 2870 \]

### Step 4: Calculate the sum of squares from \(1^2\) to \(4^2\)

Next, we need to subtract the sum of squares from \(1^2\) to \(4^2\):

\[ 1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30 \]

### Step 5: Subtract from the total sum

Now, we subtract this from the total sum we calculated earlier:

\[ \text{Sum from } 5^2 \text{ to } 20^2 = 2870 - 30 = 2840 \]

### Conclusion

Thus, the sum of the series \(5^2 + 6^2 + 7^2 + \ldots + 20^2\) is:

\[ \boxed{2840} \]