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.
Hi
You can see it works for $\displaystyle n=1,$ i.e. $\displaystyle \sum\limits_{r=1}^1r^3=\left( \frac{1(1+1)}{2}\right)^2$
Now assume it is true for some $\displaystyle n\geq 0.$
Write:
$\displaystyle \sum\limits_{r=1}^{n=1}r^3$
$\displaystyle =\sum\limits_{r=1}^nr^3+(n+1)^3$
$\displaystyle =\left( \frac{n(n+1)}{2}\right)^2+(n+1)^3$ (by induction hypothesis)
$\displaystyle =\left( \frac{n(n+1)}{2}\right)^2+(n+1)^2(n+1)$
Can you end?
Well the idea is to factorize by $\displaystyle (n+1)^2$ which appears in the two terms:
$\displaystyle \left( \frac{n(n+1)}{2}\right)^2+(n+1)^2(n+1)$
$\displaystyle =(n+1)^2\left( \frac{n^2}{4}+(n+1)\right)$
$\displaystyle =(n+1)^2\left( \frac{n^2+4n+4}{4}\right)$
$\displaystyle =(n+1)^2\left( \frac{(n+2)^2}{4}\right)$
$\displaystyle =\left( \frac{(n+1)(n+2)}{2}\right)^2$
which is what you wanted.