Question

If $n>1,x>-1,x\ne 0$, then the statement ${(1+x)}^n>1+nx$ is true for

• A. $nϵN$
• B. $\forall n>1$
• C. $x>-1andx\ne 0$
• D. None of these

Answer 1

The correct option is A $nϵN$

$P\left(1\right)$ is not true

For $n=2,P\left(2\right):{(1+x)}^2>1+2x$ is true if $x\ne 0$

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

$\therefore {(1+x)}^{k+1}=(1+x){(1+x)}^k>(1+x)(1+kx)>1+(k+1)x+k{x}^2>1+(k+1)x$

$(\because k{x}^2>0)$

$\therefore$ By PMI

Given statement is true for every $n\in N$.