## Algebra 2 (1st Edition)

Proofs using mathematical induction consist of two steps: 1) The base case: here we prove that the statement holds for the first natural number. 2) The inductive step: assume the statement is true for some arbitrary natural number $n$ larger than the first natural number, then prove that the statement also holds for $n + 1$. Hence here: 1) For $n=1: (1)(1+1)(2(1)+1)/6=1^2$. 2) Assume for $n=k: 1+4+...+k^2=\frac{k(k+1)(2k+1)}{6}$. Then for $n=k+1$: $1+4+...+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+k^2+2k+1=\frac{(k+1)(k+2)(2(k+1)+1)}{6}.$ Thus we proved what we wanted to.