<s>[INST] How to find the sum of: $a^2+(a+d)^2+(a+2d)^2+(a+3d)^2+\cdots+[a+(n-1)d]^2?$ Given the arithmetic series with a common difference, d, first term, a, and n is the nth terms.
Deriving the formula for $S_n$
$$S_n=a+(a+d)+(a+2d)+(a+3d)+\cdots+[a+(n-1)d]\tag1$$
Reverse the sum
$$S_n=[a+(n-1)d]+\cdots+(a+d)+a\tag2$$
$(1)+(2)$
$$2S_n=n[a+(n-1)d]$$
$$S_n={n\over 2}[a+(n-1)d]$$

How do we go about to find the square sum of the arithmetic series?

$$T_n=a^2+(a+d)^2+(a+2d)^2+(a+3d)^2+\cdots+[a+(n-1)d]^2\tag3$$
 [/INST] The only method that I can think of to do it is by using the results that
$$
  1 + 2 + \dots + n = \frac{n(n + 1)}{2}
$$
and
$$
  1^2 + 2^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6}.
$$
We then have that
$$ \begin{align}
  T_n & = a^2 + (a + d)^2 + \dots + (a + (n - 1)d)^2 \\
 & = a^2 + (a^2 + 2ad + d^2) + (a^2 + 4ad + 4d^2) + \dots + (a^2 + 2(n - 1)ad + (n - 1)^2 d^2) \\
 & = na^2 + 2ad(1 + 2 + \dots + (n - 1)) + d^2 (1^2 + 2^2 + \dots + (n - 1)^2) \\
 & = na^2 + 2ad \frac{(n-1)n}{2} + d^2 \frac{(n - 1)n(2n - 1)}{6} \\
 & = na^2 = n(n - 1) ad + \frac{1}{6} n(n - 1)(2n - 1) d^2.
\end{align} $$

The easiest way to find the formula for the sum
$$
  1^2 + 2^2 + \dots + n^2
$$
is by considering
$$
  (n + 1)^3 - n^3 = 3n^2 + 3n + 1.
$$
We then have that
$$ \begin{align}
  (n + 1)^3 & = \left( (n + 1)^3 - n^3 \right) + \left( n^3 - (n - 1)^3 \right) + \dots + \left( 1^3 - 0^3 \right) \\
 & = (3n^2 + 3n + 1) + (3(n - 1)^3 + 3(n - 1) + 1) + \dots + (3 \cdot 0^3 + 3 \cdot 0 + 1) \\
 & = 3(0^2 + 1^2 + 2^2 + \dots + n^2) + 3(0 + 1 + 2 + \dots + n) + (1 + 1 + \dots + 1) \\
 & = 3(1^2 + 2^2 + \dots + n^2) + 3 \frac{n(n + 1)}{2} + (n + 1).
\end{align} $$
It follows that
$$
  1^2 + 2^2 + \dots + n^2 = \frac{1}{3} (n + 1)^3 - \frac{1}{2} n(n + 1) - \frac{1}{3} (n + 1) = \frac{n + 1}{6} \left(2(n + 1)^2 - 3 n - 2 \right) = \frac{(n + 1) (2n^2 + n)}{6} = \frac{n(n + 1)(2n + 1)}{6}. 
$$
Alternatively, if someone told you what the formula is, you could prove that it is correct by induction.
</s>