Answer
See the explanation below.
Work Step by Step
Let us assume ${{S}_{n}}$ be a statement involving the positive integer n. If
1. ${{S}_{1}}$ is true, and
2. Let us assume that the ${{S}_{k}}$ be true for some integer $ k $.
Then, one has to use ${{S}_{k}}$ to prove ${{S}_{k+1}}$ is also true. If one is able to prove ${{S}_{k+1}}$ then by the principle of mathematical induction the statement ${{S}_{n}}$ is true for all positive integers $ n $.
Firstly, show that ${{S}_{1}}$ is true.
For ${{S}_{1}}$ one has
And write ${{S}_{1}}$ by taking its term on left and substituting n with $1$ on the right.
${{S}_{1}}:\sum\limits_{i=1}^{1}{5\cdot {{6}^{i}}=6\left( {{6}^{1}}-1 \right)}$
So, $5\cdot 6=5\cdot 6$
Therefore, the statement ${{S}_{1}}$ is true.
Then, assume that ${{S}_{k}}$ is true, then ${{S}_{k+1}}$ should also be true by the method of mathematical induction.
${{S}_{k}}:\sum\limits_{i=1}^{k}{5\cdot {{6}^{i}}=6\left( {{6}^{k}}-1 \right)}$
Now, one has to prove ${{S}_{k+1}}$ is true
${{S}_{k+1}}:\sum\limits_{i=1}^{k+1}{5\cdot {{6}^{i}}=6\left( {{6}^{k+1}}-1 \right)}$
Since, one assume that ${{S}_{k}}$ is true,
$\sum\limits_{i=1}^{k}{5\cdot {{6}^{i}}=6\left( {{6}^{k}}-1 \right)}$
By adding by $5\cdot {{6}^{k+1}}$ on both sides,
$\begin{align}
& \sum\limits_{i=1}^{k}{5\cdot {{6}^{i}}+\left( 5\cdot {{6}^{k+1}} \right)=6\left( {{6}^{k}}-1 \right)}+\left( 5\cdot {{6}^{k+1}} \right) \\
& \sum\limits_{i=1}^{k+1}{5\cdot {{6}^{i}}=6\left( {{6}^{k}}-1 \right)}+\left( \left( 6-1 \right)\cdot {{6}^{k+1}} \right) \\
& \sum\limits_{i=1}^{k+1}{5\cdot {{6}^{i}}=\left( {{6}^{k+1}}-6 \right)}+\left( {{6}^{k+2}}-{{6}^{k+1}} \right) \\
& \sum\limits_{i=1}^{k+1}{5\cdot {{6}^{i}}=}\left( {{6}^{k+2}}-6 \right)
\end{align}$
So,
$\sum\limits_{i=1}^{k+1}{5\cdot {{6}^{i}}=6\left( {{6}^{k+1}}-1 \right)}$
So, the final statement is ${{S}_{k+1}}$.
Thus, by the method of principal of mathematical induction, the statement $\sum\limits_{i=1}^{n}{5\cdot {{6}^{i}}=6\left( {{6}^{n}}-1 \right)}$ is true for every positive integer $ n $.
