Answer
See below.
Work Step by Step
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 $n=1: 4=0.5(1)(9-1)$.
2) Assume for $n=k: 4+3+...+(5-k)=0.5k(9-k)$. Then for $n=k+1$:
$4+3+...+(5-k)+(4-k)=0.5k(9-k)+(4-k)=0.5k^2+4.5k+4-k=0.5(k+1)(8-k)=0.5(k+1)(9-(k+1)).$
Thus we proved what we wanted to.