## College Algebra (6th Edition)

By substituting in: for $n=3$: $3+7+11=3(6+1)$ for $n=k+1$, before simplication: $3+7+11+...+(4(k+1)-1)=(k+1)(2(k+1)+1)$, which is equivalent to: $3+7+11+...+(4k+3)=(k+1)(2k+3)$,