Let's break this problem into two parts:

1. Find the algebraic form of the sum of the sequence.
2. Prove that the sum of any terms of this arithmetic sequence, starting from the first, added to one gives a perfect square.

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Part 1: Finding the Algebraic Form of the Sequence

The sequence provided is:

$15, 33, 51, \dots$

This is an arithmetic sequence, where the first term $a_1 = 15$ and the common difference $d = 33 - 15 = 18$.

The general formula for the $n$-th term of an arithmetic sequence is:

$a_n = a_1 + (n-1) \cdot d$

Substitute the known values:

$a_n = 15 + (n-1) \cdot 18$

Simplify the equation:

$a_n = 15 + 18n - 18 = 18n - 3$

Thus, the algebraic form for the $n$-th term of the sequence is:

$a_n = 18n - 3$

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Part 2: Proving that the Sum of Any Terms Added to One Forms a Perfect Square

We are tasked with proving that for any integer $n$, the sum of the first $n$ terms of this arithmetic sequence, when added to one, is a perfect square.

Step 1: Sum of the First $n$ Terms of the Arithmetic Sequence

The sum of the first $n$ terms $S_n$ of an arithmetic sequence is given by the formula:

$S_n = \frac{n}{2} \cdot (a_1 + a_n)$

From earlier, we know that:

[ a_1 = 15 \quad \text{and} \quad a_n