Prove that $1^2 + 2^2 + ..... + (n-1)^2 < \frac {n^3} { 3} < 1^2 + 2^2 + ...... + n^2$

Question

I'm having trouble on starting this induction problem.

The question simply reads : prove the following using induction:

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

Answer 1

\begin{align} 1^2+2^2+...+&(n-1)^2 \quad\quad= \frac{n(n-1)(2n)}{6} = \frac{n^2(n-1)}{3} &< \frac{n^3}{3} \\ 1^2+2^2+...+&(n-1)^2+n^2 =\frac{n(n+1)(2n+1)}{6} =\frac{2n^3+3n^2+n}{6} =\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6} &>\frac{n^3}{3} \end{align}

Cleary our inequality is proven.

Answer 2

If you wish a more direct application of induction, we have that if

$$ 1^2+2^2+\cdots+(n-1)^2 < \frac{n^3}{3} $$

then

$$ 1^2+2^2+\cdots+n^2 < \frac{n^3+3n^2}{3} < \frac{n^3+3n^2+3n+1}{3} = \frac{(n+1)^3}{3} $$

Similarly, if

$$ 1^2+2^2+\cdots+n^2 > \frac{n^3}{3} $$

then

$$ 1^2+2^2+\cdots+(n+1)^2 > \frac{n^3+3n^2+6n+3}{3} > \frac{n^3+3n^2+3n+1}{3} = \frac{(n+1)^3}{3} $$

Answer 3

$$0^2 + 1^2 + \ldots + (n - 1)^2 \leqslant \int_0^nx^2\,dx \leqslant 1^2 + 2^2 + \ldots + n^2,$$

and it's quite obvious that current integral is exactly $\frac{n^3}{3}$.

Answer 4

I would start by proving each side of the equation using induction.

First prove that $1^2+2^2+..+n^2 = \sum\limits_{i=1}^n i^2 \geq \frac{n^3}{3}$

Induction says that we must first check that the relationship holds for $n=1$.

$\sum\limits_{i=1}^1 i^2 = 1 \geq \frac{1^3}{3}=\frac{1}{3}$

So it holds for $n=1$. Now we assume that it is true that $\sum\limits_{i=1}^n i^2 \geq \frac{n^3}{3}$ for $n=k$, and prove that the result can be obtained from $n=k+1$.

$\sum\limits_{i=1}^{k+1} i^2 = \sum\limits_{i=1}^k i^2 + (k+1)^2$

$\sum\limits_{i=1}^{k+1} i^2 \geq \frac{k^3}{3} + k^2+2k+1$

$\sum\limits_{i=1}^{k+1} i^2 \geq \frac{k^3+3k^2+3k+1}{3}+k+\frac{2}{3}$

$\sum\limits_{i=1}^{k+1} i^2 \geq \frac{(k+1)^3}{3}+k+\frac{2}{3}$

$\sum\limits_{i=1}^{k+1} i^2 \geq \frac{(k+1)^3}{3}$

Since $k+\frac{2}{3}$ is a positive number, removing it from the right side still maintains the inequality, and the first part of your question has been proved by induction. Do a similar thing to the left hand side to prove it.

Answer 5

Perhaps simpler, the induction step is done if you show $$3n^2<(n+1)^3-n^3<3(n+1)^2$$

But $(n+1)^3-n^3=3n^2+3n+1$ which is clearly in between $3n^2$ and $3(n+1)^2$.