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

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.

Tunk-Fey
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Ali Mustafa
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    After getting rid of the obfuscating $n^3$ in the denominator (it's on both sides and can be cancelled), apply induction. It's not that hard. You can do it yourself (if you've seen induction); come back if you get stuck. Much more satisfying. :) – gnometorule Jul 06 '14 at 07:32

3 Answers3

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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.

Community
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Peter Woolfitt
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  • You wrote (n+1)(2n+1), but in my formula I wrote (n−1)(2n−1). Was that a mistake or am I missing something? – Ali Mustafa Jul 06 '14 at 07:42
  • Just take a look at the upper boundary of the summation and everything should be clear – b00n heT Jul 06 '14 at 07:43
  • @AliMustafa boon heT is correct, here I am summing to $n$, and in your summation you are summing to $n-1$. Letting $n=k-1$ in the equation above yields what you want, namely $\dfrac{(k-1)k(2k-1)}{6}$. – Peter Woolfitt Jul 06 '14 at 07:49
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Let $S=\Sigma i^2=1^2+2^2+3^2+...+n^2$
Now, $r^3-(r-1)^3=3r^2-3r+1$ So $$\Sigma_0^n \left[r^3-(r-1)^3\right]=3\Sigma_0^nr^2-3\Sigma_0^nr+\Sigma_0^n1\\ =3S-3\frac{n(n+1)}2+n\\ \implies 3S=n^3-n+\frac{3n(n+1)}2\\ \large S=\frac{n(n+1)(2n+1)}6$$

RE60K
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0

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 $$ \large\frac{1}{n^3}\sum_{k=1}^{n-1} k^2=\color{blue}{\frac{1^2+2^2+\cdots+(n-1)^2}{n^3}=\frac{n(n-1)(2n-1)}{6n^3}}. $$

Tunk-Fey
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