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On page 81 of the 1st volume he writes the following: $$ (1^2+2^2+...+(n-1)^2)/n^3 = (n-1)n(2n-1)/6n^3 $$ Can someone explain how the left side of the equation can be transformed into the expression on the right side. I don't understand the authors explanation, which says the following:

If in the obvious equalities $(k+1)^3-k^3=3k^2+3k+1$, for the different values $k=1,2,\ldots,n-1$, we add the left and the right sides separately, we obtain the equation $n^3-1=3\rho+3(n-1)n/2+n-1$, where $\rho=1^2+2^2+\ldots+(n-1)^2$. Solving this equation for $\rho$, we get $\rho=(n-1)n(2n-1)/6$

Vincenzo Tibullo
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Ali Mustafa
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  • have a look at the proofs here: http://math.stackexchange.com/questions/320985/how-to-determine-equation-for-sum-k-1n-k3/ The one you need s very similar – Alex Jul 05 '14 at 08:20

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$$(n+1)^3-1=\sum_{k=1}^n(k+1)^3-k^3=\sum_{k=1}^n(3\color{green}{k^2}+3\color{red}{k}+\color{blue}{1})=3\color{green}{\sum_{k=1}^nk^2}+3\color{red}{\sum_{k=1}^nk}+\color{blue}{\sum_{k=1}^n1}$$ If one can identify the two last sums $\color{red}{\sum\limits_{k=1}^nk}$ and $\color{blue}{\sum\limits_{k=1}^n1}$ on the RHS, one gets the first sum $\color{green}{\sum\limits_{k=0}^nk^2}$ on the RHS. Thus, $$ 3\color{green}{\sum_{k=1}^nk^2}=(n+1)^3-1-3\cdot\color{red}{\frac{n(n+1)}2}-\color{blue}{n}=\ldots$$

Did
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