Finding the Value of the Series of Even Number Squares

The problem asks us to find the value of the series \( (2^2 + 4^2 + 6^2 + \text{...} + 22^2) \). This is a series where each term is the square of an even number.

Let's examine the terms in the series:

• The first term is \(2^2 = (2 \times 1)^2\)
• The second term is \(4^2 = (2 \times 2)^2\)
• The third term is \(6^2 = (2 \times 3)^2\)
• ...
• The last term is \(22^2\). To find which term this is, we can write \(22\) as \(2 \times n\). So, \(2n = 22\), which means \(n = 11\). The last term is \( (2 \times 11)^2 \).

So, the series can be written as the sum of terms of the form \( (2n)^2 \) for \(n\) from 1 to 11.

The series is \( \sum_{n=1}^{11} (2n)^2 \).

We can simplify the general term \( (2n)^2 \):

\( (2n)^2 = 2^2 \times n^2 = 4n^2 \)

So, the sum is \( \sum_{n=1}^{11} 4n^2 \).

Using the property of summation that allows factoring out a constant, we get:

\( \sum_{n=1}^{11} 4n^2 = 4 \sum_{n=1}^{11} n^2 \)

Now we need to find the sum of the first 11 squares, which is \( \sum_{n=1}^{11} n^2 \). The formula for the sum of the first \(k\) squares is given by:

\( \sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6} \)

In this case, \(k = 11\). Substituting \(k=11\) into the formula:

\( \sum_{n=1}^{11} n^2 = \frac{11(11+1)(2 \times 11+1)}{6} \)

\( \sum_{n=1}^{11} n^2 = \frac{11(12)(22+1)}{6} \)

\( \sum_{n=1}^{11} n^2 = \frac{11 \times 12 \times 23}{6} \)

Now, let's perform the calculation:

\( \frac{11 \times 12 \times 23}{6} = 11 \times \frac{12}{6} \times 23 \)

\( = 11 \times 2 \times 23 \)

\( = 22 \times 23 \)

Calculating \(22 \times 23\):

\( 22 \times 23 = 22 \times (20 + 3) = 22 \times 20 + 22 \times 3 = 440 + 66 = 506 \)

So, the sum of the first 11 squares is \( \sum_{n=1}^{11} n^2 = 506 \).

Now we need to find the total sum of the given series, which is \( 4 \sum_{n=1}^{11} n^2 \).

Total sum = \( 4 \times 506 \)

\( 4 \times 506 = 4 \times (500 + 6) = 4 \times 500 + 4 \times 6 = 2000 + 24 = 2024 \)

The value of the series \( (2^2 + 4^2 + 6^2 + \text{...} + 22^2) \) is 2024.

Understanding Series Summation and Formulas

This problem involves summing a specific type of series. Recognizing the pattern and using known summation formulas is key.

• The series consists of squares of consecutive even numbers.
• Each even number can be written in the form \(2n\).
• The series runs up to \(22 = 2 \times 11\), indicating the index \(n\) goes from 1 to 11.
• The sum of squares formula is a standard result in arithmetic progressions and series.

Steps to Calculate the Series Sum

1. Identify the pattern of the terms in the series.
2. Write the general term of the series using an index (e.g., \(n\)).
3. Determine the range of the index (from where to where the sum runs).
4. Rewrite the sum using summation notation.
5. Simplify the general term if possible and factor out constants.
6. Use known summation formulas (like the sum of first \(n\) squares) if applicable.
7. Substitute the upper limit of the index into the formula.
8. Perform the necessary calculations to find the final sum.

Revision Table: Key Formulas

Additional Information on Series and Summation

A series is the sum of the terms of a sequence. Understanding different types of sequences (like arithmetic progression, geometric progression) and their corresponding series is important. Summation notation (\( \Sigma \)) is a compact way to represent the sum of a series.

Properties of summation:

• Constant multiple: \( \sum_{n=a}^{b} c \cdot f(n) = c \sum_{n=a}^{b} f(n) \)
• Sum/Difference: \( \sum_{n=a}^{b} (f(n) \pm g(n)) = \sum_{n=a}^{b} f(n) \pm \sum_{n=a}^{b} g(n) \)
• Additivity: \( \sum_{n=a}^{c} f(n) + \sum_{n=c+1}^{b} f(n) = \sum_{n=a}^{b} f(n) \) (where \(a \le c < b\))

For the series of squares of even numbers, we used the constant multiple property to factor out 4, simplifying the problem to finding the sum of consecutive squares.