I have an assignment question
Prove by either the Well Ordering Principle or induction that for all nonnegative integers $n$: $$\sum_{k=0}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2.$$
I am able to solve this question using basic Induction, but not able to figure out how to do it by using Well Ordering Principle.
Any Solutions or hints would be very helpful. Thanks in Advance
On
For $n=1$ the left side is equal to $1$ and the right side is also equal to $1$ so the equality holds true. Suppose that the equality $(1)$ holds true for some $n\geq1$ then for $n+1$ we obtain $$\sum_{i=1}^{n+1} i^3 =\sum_{i=1}^{n} i^3 +(n+1)^3 = \frac{n^2 (n+1)^2 }{4} +(n+1)^3 =(n+1)^2\left(\frac{1}{4} n^2 +n+1\right) =(n+1)^2 \left(\frac{1}{2} n +1\right)^2 =\left(\frac{(n+1)(n+2)}{2}\right)^2 .\fbox{}$$
Suppose that the statement is not true.
Then the set $A=\left\{ n\in\mathbb{N}\mid\sum_{i=0}^{n}i^{3}\neq\left(\frac{n\left(n+1\right)}{2}\right)^{2}\right\} $ is not empty.
Since $\mathbb{N}$ is well-ordered set $A$ has a minimal element $m$.
That means that $\sum_{i=0}^{n}i^{3}=\left(\frac{n\left(n+1\right)}{2}\right)^{2}$ is true for $n<m$ and is not true for $n=m$.
From this you can deduce a contradiction.
(Start with $\sum_{i=0}^{m-1}i^{3}=\left(\frac{\left(m-1\right)m}{2}\right)^{2}$ and prove on base of that $\sum_{i=0}^{m}i^{3}=\sum_{i=0}^{m-1}i^{3}+m^{3}=\left(\frac{m\left(m+1\right)}{2}\right)^{2}$)
The conclusion is then that $A=\emptyset$ wich is exactly the statement to be proven.