## College Algebra (6th Edition)

Proofs using mathematical induction consists 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 we prove that then the statement also holds for $n + 1$. Hence here: 1) For $n=1: 4=2(1)(1+1)=4$. 2) Assume for $n=k: 4+8+...+4k=2k(k+1)$. Then for $n=k+1$: $4+8+...+4k+4k+4=2k(k+1)+4k+4=2k^2+2k+4k+4=2(k+1)(k+2)=2(k+1)((k+1)+1).$ Thus we proved what we wanted to.