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From the two equations below:

$1^{2}+2^{2}+...+n^{2} = {n^3 \over 3}+{n^2 \over 2}+{\frac n6}$

$1^{2}+2^{2}+...+(n-1)^{2} = {n^3 \over 3}-{n^2 \over 2}+{\frac n6}$

How can the following inequalities be deduced:

$1^{2}+2^{2}+...+(n-1)^{2} < {n^{3} \over 3} < 1^{2}+2^{2}+...+n^{2}$ for $n\ge 1$

I understand that it can be proved via induction, but how can we deduce the answer before we test that it is correct via induction.

This is part of a proof from the introduction of Tom M Apostol's book Calculus 1.

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From the two equations you wrote, you have to prove that :

${n^2 \over 2}+{\frac n6} > 0$

$-{n^2 \over 2}+{\frac n6} < 0$

to get to the inequality you want. These inequalities are both true for $n\ge 1$

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We don't even need the explicit formula. If we set $a_n=\sum_{k=1}^{n}k^2$, we may prove $$ a_{n-1} < \frac{n^3}{3} < a_n $$ by induction. That is obviously true for $n=1$, and $$ \frac{(n+1)^3}{3}-\frac{n^3}{3} = n^2+n+\frac{1}{3} $$ is clearly between $n^2$ and $(n+1)^2$.

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  • $\begingroup$ Thank you Jack. Sorry I should have been clearer in my question. How can you initially deduce that (n^3)/3 is the correct answer? $\endgroup$ – Jamie Fearon Nov 4 '15 at 15:14
  • $\begingroup$ @JamieFearon: the correct answer to which question? $\sum_{k=1}^{n}k^2$ has to behave like $\frac{n^3}{3}$ since $\int_{0}^{n}x^2\,dx = \frac{n^3}{3}.$ $\endgroup$ – Jack D'Aurizio Nov 4 '15 at 15:21

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