Can you show that the LHS equals the RHS in this equation, by showing how I can get the expression on the RHS?

Question

$$ \frac{1^2+2^2+...+(n-1)^2}{n^3} = \frac{(n-1)n(2n-1)}{6n^3} $$

Can someone show me step by step how I can transform the LHS to the RHS? If possible, using high school-level math.

I have now edited the title, because I didn't formulate myself right. I would like to know how I can derive the expression on the RHS, without knowing it.

Answer 1

Hint: you can prove by induction that $$\sum\limits_{i=1}^ni^2=\frac{n(n+1)(2n+1)}{6}$$

There are also many other ways to prove this. See this Math.SE post.

Answer 2

Let $S=\Sigma i^2=1^2+2^2+3^2+...+n^2$
Now, $r^3-(r-1)^3=3r^2-3r+1$ So $$\Sigma_0^n \left[r^3-(r-1)^3\right]=3\Sigma_0^nr^2-3\Sigma_0^nr+\Sigma_0^n1\\ =3S-3\frac{n(n+1)}2+n\\ \implies 3S=n^3-n+\frac{3n(n+1)}2\\ \large S=\frac{n(n+1)(2n+1)}6$$

Answer 3

Consider the geometric progression $$ \sum_{k=1}^{n-1} x^k=\frac{x^{n}-x}{x-1}.\tag1 $$ Differentiating $(1)$ with respect to $x$ yields $$ \sum_{k=1}^{n-1} kx^{k-1}=\frac{(n-1)x^{n}-nx^{n-1}+1}{(x-1)^2}.\tag2 $$ Multiplying $(2)$ by $x$ yields $$ \sum_{k=1}^{n-1} kx^{k}=\frac{(n-1)x^{n+1}-nx^{n}+x}{(x-1)^2}.\tag3 $$ Differentiating $(3)$ with respect to $x$ yields $$ \sum_{k=1}^{n-1} k^2x^{k-1}=\frac{(-2n^2+2n+1)x^{n}+n^2x^{n-1}+(n-1)^2x^{n+1}-x-1}{(x-1)^3}.\tag4 $$ Setting $x=1$ to $(4)$ and evaluating the RHS at $x\to1$ by L'Hospital's rule yields \begin{align} \sum_{k=1}^{n-1} k^2&=\lim_{x\to1}\frac{(-2n^2+2n+1)x^{n}+n^2x^{n-1}+(n-1)^2x^{n+1}-x-1}{(x-1)^3}\\ &=\large\color{darkgreen}{\frac{n(n-1)(2n-1)}{6}}.\tag5 \end{align} Hence, using $(5)$ we obtain $$ \large\frac{1}{n^3}\sum_{k=1}^{n-1} k^2=\color{blue}{\frac{1^2+2^2+\cdots+(n-1)^2}{n^3}=\frac{n(n-1)(2n-1)}{6n^3}}. $$