Solving the Sum of Squares Series Problem

The problem asks us to find the sum of a specific series, given the sum of the first 14 squares. We are given that the sum of the squares of the first 14 natural numbers is 1015.

The given information is:

\(1^2 + 2^2 + 3^2 + \dots + 14^2 = 1015\)

We need to find the value of the following series:

\(3^2 + 6^2 + 9^2 + \dots + 42^2\)

Analyzing the Second Series

Let's look closely at the terms in the second series. Each term is a square of a number that is a multiple of 3:

• The first term is \(3^2\), which is \((3 \times 1)^2\).
• The second term is \(6^2\), which is \((3 \times 2)^2\).
• The third term is \(9^2\), which is \((3 \times 3)^2\).
• This pattern continues until the last term, which is \(42^2\). We can write 42 as \(3 \times 14\), so the last term is \((3 \times 14)^2\).

So, the second series can be written as:

\((3 \times 1)^2 + (3 \times 2)^2 + (3 \times 3)^2 + \dots + (3 \times 14)^2\)

Using the Property of Squares

We know that for any numbers \(a\) and \(b\), \((ab)^2 = a^2 b^2\). We can apply this property to each term in the second series:

• \((3 \times 1)^2 = 3^2 \times 1^2\)
• \((3 \times 2)^2 = 3^2 \times 2^2\)
• \((3 \times 3)^2 = 3^2 \times 3^2\)
• ...
• \((3 \times 14)^2 = 3^2 \times 14^2\)

Substituting these back into the series, we get:

\(3^2 \times 1^2 + 3^2 \times 2^2 + 3^2 \times 3^2 + \dots + 3^2 \times 14^2\)

Factoring and Calculating the Sum

Notice that \(3^2\) is a common factor in all the terms of this series. We can factor out \(3^2\):

\(3^2 (1^2 + 2^2 + 3^2 + \dots + 14^2)\)

We are given that the sum inside the parentheses is 1015. So, the expression becomes:

\(3^2 \times 1015\)

Now, we calculate the value of \(3^2\) and then multiply:

\(3^2 = 9\)

So the sum is:

\(9 \times 1015\)

Let's perform the multiplication:

The sum of the series \(3^2 + 6^2 + 9^2 + \dots + 42^2\) is 9135.

Result and Option Matching

The calculated sum is 9135.

Comparing this result with the given options:

• Option 1: 9135
• Option 2: 9325
• Option 3: 9315
• Option 4: 9235

Our calculated value matches Option 1.

Revision Table: Key Concepts

Additional Information: Sum of Squares Formula

While we used a given sum in this problem, it's useful to know the general formula for the sum of the squares of the first \(n\) natural numbers:

\(\sum_{i=1}^{n} i^2 = 1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}\)

In our problem, \(n = 14\). We could verify the given sum:

\(\frac{14(14+1)(2 \times 14+1)}{6} = \frac{14 \times 15 \times (28+1)}{6} = \frac{14 \times 15 \times 29}{6}\)

We can simplify this:

\(\frac{14}{2} \times \frac{15}{3} \times 29 = 7 \times 5 \times 29 = 35 \times 29\)

\(35 \times 29 = 35 \times (30 - 1) = 35 \times 30 - 35 \times 1 = 1050 - 35 = 1015\)

This calculation confirms the given sum is correct. Our approach, however, did not require using the formula directly, but rather the relationship between the two series based on the factoring property.