#### 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.
And write ${{S}_{1}}$ by taking the its term on left and substituting n with $1$ on the right.
${{S}_{1}}:n+2>n $
$\begin{align}
& 1+2>1 \\
& 3>1
\end{align}$
This is true.
Therefore, the statement ${{S}_{1}}$ is true.
Then, let us assume that ${{S}_{k}}$ is true, then ${{S}_{k+1}}$ should also be true by the method of mathematical induction.
${{S}_{k}}:k+2>k $
Now, one has to prove ${{S}_{k+1}}$ is true
${{S}_{k+1}}:k+1+2>k+1$
$ k+3>k+1$
Since, one assume that ${{S}_{k}}$ is true,
$ k+2>k $
By adding 1 on both sides, one get
$\begin{align}
& k+2+1>k+1 \\
& k+3>k+1
\end{align}$
So, the final statement is ${{S}_{k+1}}$.
Thus, by the principal method of of mathematical induction, the statement $ n+2>n $ is true for every positive integer $ n $.