What is the sum of the first 
n
n terms of this sequence?

The sequence we have is \(1, 4, 9, 16, 25, \ldots\), which can be expressed in terms of the general term \(a_n = n^2\).

To find the sum of the first \(n\) terms of this sequence, we need to calculate the sum of squares of the first \(n\) natural numbers. The formula for the sum of squares of the first \(n\) natural numbers is given by:

\[
S_n = 1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}
\]

### Derivation of the Formula

The formula \(\frac{n(n+1)(2n+1)}{6}\) can be derived using mathematical induction or by summing the squares directly, but for now, let’s use this formula to calculate the sum:

- **Example Calculation:**
  - If \(n = 1\), \(S_1 = \frac{1(1+1)(2 \cdot 1+1)}{6} = \frac{1 \times 2 \times 3}{6} = 1\)
  - If \(n = 

Learn how to calculate the sum of squares for the sequence 1, 4, 9, 16, 25, ... This problem involves finding the sum of squares using the formula S_n = n(n+1)(2n+1)/6. Suitable for advanced high school students and beyond.

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