1. ## Mathematical induction

Use the principle of mathematical induction to prove the
n
Σ r^3 = 1/4n^2(n+1)^2.
r=1

I need help on this problem.

2. Hi

You can see it works for $n=1,$ i.e. $\sum\limits_{r=1}^1r^3=\left( \frac{1(1+1)}{2}\right)^2$

Now assume it is true for some $n\geq 0.$

Write:

$\sum\limits_{r=1}^{n=1}r^3$
$=\sum\limits_{r=1}^nr^3+(n+1)^3$
$=\left( \frac{n(n+1)}{2}\right)^2+(n+1)^3$ (by induction hypothesis)
$=\left( \frac{n(n+1)}{2}\right)^2+(n+1)^2(n+1)$

Can you end?

3. Thank you, but I cant end till I see. Once I see it I can solve another problem. I would appreciate if ou can end it.

4. Well the idea is to factorize by $(n+1)^2$ which appears in the two terms:

$\left( \frac{n(n+1)}{2}\right)^2+(n+1)^2(n+1)$
$=(n+1)^2\left( \frac{n^2}{4}+(n+1)\right)$
$=(n+1)^2\left( \frac{n^2+4n+4}{4}\right)$
$=(n+1)^2\left( \frac{(n+2)^2}{4}\right)$
$=\left( \frac{(n+1)(n+2)}{2}\right)^2$

which is what you wanted.

5. Originally Posted by clic-clac
Hi

You can see it works for $n=1,$ i.e. $\sum\limits_{r=1}^1r^3=\left( \frac{1(1+1)}{2}\right)^2$

Now assume it is true for some $n\geq 0.$

Write:

$\sum\limits_{r=1}^{n=1}r^3$
$=\sum\limits_{r=1}^nr^3+(n+1)^3$
$=\left( \frac{n(n+1)}{2}\right)^2+(n+1)^3$ (by induction hypothesis)
$=\left( \frac{n(n+1)}{2}\right)^2+(n+1)^2(n+1)$

Can you end?
i think he should assume that it is true for $n\geq 1$,no ?

6. Yeah indeed it would be better according to what was done before

By the way the case $n=0$ is also true, but isn't very interesting (nothing more than $0=0$)