Find the sum to n terms of the series :`5^2+6^2+7^2+dotdotdot+20^2`

Explanation

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

### Step 1: Identify the series
The series we need to sum is:
\[ 5^2 + 6^2 + 7^2 + \ldots + 20^2 \]

### Step 2: Use the formula for the sum of squares
The sum of squares of the first \(n\) natural numbers is given by the formula:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]

### Step 3: Calculate the sum from 1 to 20
We first calculate the sum of squares from \(1^2\) to \(20^2\):
\[ \sum_{k=1}^{20} k^2 = \frac{20(20+1)(2 \cdot 20 + 1)}{6} \]
Calculating this:
\[ = \frac{20 \cdot 21 \cdot 41}{6} \]

### Step 4: Calculate the sum from 1 to 4
Next, we calculate the sum of squares from \(1^2\) to \(4^2\):
\[ \sum_{k=1}^{4} k^2 = \frac{4(4+1)(2 \cdot 4 + 1)}{6} \]
Calculating this:
\[ = \frac{4 \cdot 5 \cdot 9}{6} \]

### Step 5: Combine the results
Now, we can find the sum from \(5^2\) to \(20^2\) by subtracting the sum from \(1^2\) to \(4^2\) from the sum from \(1^2\) to \(20^2\):
\[ \sum_{k=5}^{20} k^2 = \sum_{k=1}^{20} k^2 - \sum_{k=1}^{4} k^2 \]

### Step 6: Substitute and simplify
Substituting the values we calculated:
1. For \( \sum_{k=1}^{20} k^2 \):
\[ = \frac{20 \cdot 21 \cdot 41}{6} = \frac{17220}{6} = 2870 \]

2. For \( \sum_{k=1}^{4} k^2 \):
\[ = \frac{4 \cdot 5 \cdot 9}{6} = \frac{180}{6} = 30 \]

Now, substituting these into our equation:
\[ \sum_{k=5}^{20} k^2 = 2870 - 30 = 2840 \]

### Final Answer
Thus, the sum to \(n\) terms of the series \(5^2 + 6^2 + 7^2 + \ldots + 20^2\) is:
\[ \boxed{2840} \]