Proof by Mathematical Induction
The process
- Show the statement works for \(n=1\)
- Assume the statement is true for \(n=k\)
- Show if true for \(n=k\) then must be true for \(n=k+1\).
- Conclusion: If it is true for \(n=1\) and true for \(n=k\) then it is true for \(n=k+1\) making it true for all natural\(n\)
Prove by induction for all positive \(n\), that \(\sum_{r=1}^{n}r^3=\frac{1}{4}n^2(n+1)^2\)
- Sub in \(n=1\)
\(1^2=\frac{1}{4}(1)(4)=1\) - We assume equal for \(n=k\)
- \(\sum_{r=1}^{k+1}r^3=\sum_{r=1}^{k}r^3+(k+1)^3\)
\(\sum_{r=1}^{k+1}r^3=\frac{1}{4}k^2(k+1)^2+(k+1)^3\)
\(=(\frac{1}{4}k^3+\frac{1}{4}k^2)^2+(k+1)^3\)
\(=(\frac{1}{4}k^6+\frac{1}{4}k^4)+(k+1)^3\)
\(=\frac{1}{4}(k^6+k^4)+(k+1)^3\)
\(=\frac{1}{4}(k+1)^2(k+1+1)^2\)