Find the sum of the series `2^2+4^2+6^2+....+(20)^2`

Options

A. 1500
B. 1540
C. 1600
D. None of these

Explanation

To find the sum of the series \(2^2 + 4^2 + 6^2 + \ldots + 20^2\), we can follow these steps:

### Step 1: Identify the series
The series consists of the squares of the first 10 even numbers. We can express the terms of the series as:
\[ 2^2, 4^2, 6^2, \ldots, (2n)^2 \]
where \(n\) goes from 1 to 10.

### Step 2: General term of the series
The general term of the series can be written as:
\[ T_n = (2n)^2 = 4n^2 \]
for \(n = 1, 2, 3, \ldots, 10\).

### Step 3: Write the sum of the series
We need to find the sum of the series:
\[ S = \sum_{n=1}^{10} T_n = \sum_{n=1}^{10} 4n^2 \]
This can be factored out as:
\[ S = 4 \sum_{n=1}^{10} n^2 \]

### Step 4: Use the formula for the sum of squares
The formula for the sum of the squares of the first \(n\) natural numbers is:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]
For \(n = 10\):
\[ \sum_{k=1}^{10} k^2 = \frac{10(10 + 1)(2 \cdot 10 + 1)}{6} = \frac{10 \cdot 11 \cdot 21}{6} \]

### Step 5: Calculate the sum of squares
Calculating step-by-step:
1. \(10 \cdot 11 = 110\)
2. \(2 \cdot 10 + 1 = 21\)
3. \(110 \cdot 21 = 2310\)
4. Now divide by 6:
\[ \frac{2310}{6} = 385 \]

### Step 6: Substitute back into the sum
Now substitute back into the equation for \(S\):
\[ S = 4 \cdot 385 = 1540 \]

### Final Result
Thus, the sum of the series \(2^2 + 4^2 + 6^2 + \ldots + 20^2\) is:
\[ \boxed{1540} \]