Question on how the sum of Arithmetic Progression is $\frac{n^3}{3}$-$\frac{n^2}{2}$+$\frac{n}{6}$.

Question

Given a set S = $({1,2,...,(n-1))}$, we obtain the sum of all of its entries through $$S_n= \frac{1}{2}n(n-1)$$ that we derive from $S_n=\frac{1}{2}n(a_1 +a_n)$. The issue is, let us square our set, $S_n^2=({1^2,2^2,...,(n-1)^2)}$, from this we get $$S^2_n =\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}$$. That is the question, though, how do we get this value,

I am confused about how to reach that. I have tried to go through $$\frac{1}{2}(n-1)((n-1)^2+1^2)$$, the problem with this is that it gives off a result that does not equal the one mentioned in the title, so there is no point in trying to simplify it. Can someone help me? Thank you.

Answer 1

The formula for the sum of squares is given by $$1^2+2^2+\ldots+n^2=\frac16 n(n+1)(2n+1).$$ Applying this, we have $$1^2+\ldots+(n-1)^2=\frac16(n-1)n(2n-1)=\frac13 n^3-\frac12n^2+\frac16n\,.$$

Your mistake is to use the AP formula when you do not have an arithmetic progression.

Answer 2

Hint. One may notice that $$ \left[\frac{1}{3} (k+1)^3-\frac{1}{2} (k+1)^2+\frac{k+1}{6}\right]-\left[\frac{k^3}{3}-\frac{k^2}{2}+\frac{k}{6}\right]=k^2,\qquad k=1,2,...,n-1, $$ and use a telescoping series.