Given `2^2+4^2+6^2+……+40^2=11480` then the value of
`1^2+2^2+3^2+…+20^2` is

Options

A. 2870
B. 2868
C. 2867
D. 2869

Answer

To find the value of \(1^2 + 2^2 + 3^2 + \ldots + 20^2\), we can use the information given in the question regarding the sum of squares of even numbers.

### Step 1: Understand the given series
The series given is:
\[ 2^2 + 4^2 + 6^2 + \ldots + 40^2 \]
This can be expressed in terms of \(n\) where \(n\) is the number of terms. The even numbers can be represented as \(2n\) where \(n = 1, 2, 3, \ldots, 20\).

### Step 2: Identify the number of terms
The last term \(40\) corresponds to \(n = 20\). Therefore, there are \(20\) terms in this series.

### Step 3: Write the sum of squares of even numbers
The sum of squares of the first \(n\) even numbers can be calculated using the formula:
\[ \text{Sum} = 2^2 + 4^2 + 6^2 + \ldots + (2n)^2 = 4(1^2 + 2^2 + 3^2 + \ldots + n^2) \]
Thus, we can express the sum as:
\[ 2^2 + 4^2 + 6^2 + \ldots + 40^2 = 4(1^2 + 2^2 + 3^2 + \ldots + 20^2) \]

### Step 4: Set up the equation
Given that:
\[ 2^2 + 4^2 + 6^2 + \ldots + 40^2 = 11480 \]
We can substitute this into our equation:
\[ 4(1^2 + 2^2 + 3^2 + \ldots + 20^2) = 11480 \]

### Step 5: Solve for \(1^2 + 2^2 + 3^2 + \ldots + 20^2\)
Now, divide both sides by \(4\):
\[ 1^2 + 2^2 + 3^2 + \ldots + 20^2 = \frac{11480}{4} = 2870 \]

### Final Answer
Thus, the value of \(1^2 + 2^2 + 3^2 + \ldots + 20^2\) is:
\[ \boxed{2870} \]