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 $n=1: \sum_{i=1}^{1} 5\cdot6^i=5\cdot6^1=30=6(6^1-1)$
2) Assume for $n=k: \sum_{i=1}^{k} 5\cdot6^i=6(6^k-1)$.
Then for $n=k+1$: $ \sum_{i=1}^{k+1} 5\cdot6^i=(\sum_{i=1}^{k} 5\cdot6^i)+5\cdot6^{k+1}=6(6^k-1)+5\cdot6^{k+1}=6^{k+1}-6+5\cdot6^{k+1}=6\cdot6^{k+1}-6=6(6^{k+1}-1)$. Thus we proved what we wanted to.