Prove that the sum of the first n cubes is given by
1^3+2^3+3^3+4^3+.....+n^3=(n^2(n+1)^2)/4
You can prove this by using Induction.
There are two steps:
1) Try your equation with a positive integer. If its equal, get to the second step.
2) Show that its equal for f(n) and f(n+1)
$\displaystyle 1^3 + 2^3 + 3^3 + ........ + n^3 = (\frac {n(n+1)}{2})^2$
1) $\displaystyle 1^3 + 2^3 + 3^3 = (\frac {3(3+1)}{2})^2 = 36$
2) $\displaystyle f(n) = 1^3 + 2^3 + 3^3 + ........ + n^3 = (\frac {n(n+1)}{2})^2$
$\displaystyle f(n+1) = 1^3 + 2^3 + 3^3 + ........ + n^3 + (n+1)^3 = (\frac {(n+1)(n+2)}{2})^2$
If you noticed, f(n+1) includes f(n). So we will replace $\displaystyle 1^3 + 2^3 + 3^3 ....... + n^3$ with $\displaystyle (\frac {n(n+1)}{2})^2$
$\displaystyle f(n+1) = (\frac {n(n+1)}{2})^2 + (n+1)^3 = (\frac {(n+1)(n+2)}{2})^2$
$\displaystyle 0 = 0$, it means this equation is true.
What did we do there?
First we tried our formula for an integer and saw that it works.
Then, we proved that it'll work for every f(n+1). Which means its true for every n.