Let $x>-1$, then statement $P\left(n\right):(1+x{)}^n>1+nx$ is true for

Solution:

For n=2,

$P\left(2\right):(1+x{)}^2=1+2x+x^2>1+2x$,as $x\ne 0$.

Assume that

$P\left(k\right):(1+x{)}^k>1+kx$                            (1)

for some $k\in \mathbf{N},k>1$

As $x>-1$, multiplying both the sides of (1) by $1+x$, we get

$\begin{array}{l}(1+x{)}^{k+1}>(1+kx)(1+x)\\ =1+(k+1)x+k{x}^2\\ >1+(k+1)x\end{array}$

                                                                                          $\left[\because k{x}^2>0\right]$

Thus, $P(k+1)$ is true.

By the principle of mathematical induction$P\left(n\right)$ is true for
all $n>1$ provided $x\ne 0$.