Sum Series Calculation

The problem asks us to find the value of the series $3^2 + 6^2 + 9^2 + \dots + 30^2$. We are given a key piece of information: the sum of the squares of the first 10 natural numbers, $1^2 + 2^2 + 3^2 + \dots + 10^2$, equals 385.

Squares Series Relation

Let's denote the series we need to calculate as $S$.

$S = 3^2 + 6^2 + 9^2 + \dots + 30^2$

Observe that each term in this series is the square of a multiple of 3:

• $3 = 3 \times 1$
• $6 = 3 \times 2$
• $9 = 3 \times 3$
• $\dots$
• $30 = 3 \times 10$

So, we can rewrite the series $S$ as:

$S = (3 \times 1)^2 + (3 \times 2)^2 + (3 \times 3)^2 + \dots + (3 \times 10)^2$

Common Square Term Factoring

Using the algebraic rule $(a \times b)^2 = a^2 \times b^2$, we can rewrite each term:

$S = (3^2 \times 1^2) + (3^2 \times 2^2) + (3^2 \times 3^2) + \dots + (3^2 \times 10^2)$

Now, we can factor out the common term, $3^2$, from the entire series:

$S = 3^2 (1^2 + 2^2 + 3^2 + \dots + 10^2)$

Sum Result Computation

We know that $3^2 = 9$. The problem statement provides the value for the sum within the parentheses:

$1^2 + 2^2 + 3^2 + \dots + 10^2 = 385$

Substitute these values back into our expression for $S$:

$S = 9 \times (385)$

Finally, calculate the product:

$S = 3465$

Final Series Sum Answer

The calculation shows that $3^2 + 6^2 + 9^2 + \dots + 30^2 = 3465$.