Question

$(1+x{)}^n>1$ + nx for n > 2 , n $ϵ$ N
x > -1 x $\ne$ 0

Answer 1

Let P(n) be true for $n=m$
Also when $n=1,(1+x{)}^1=1+,x$. so in this case, inequality does not hold.
But when $n=2$
$\therefore (1+x{)}^2=1+2x+x^2>1+2x$
Thus P(n) is true for $n=2$
$(1+m{)}^m>1+mx$ ....(1)
Since $x>-1$, multiplying by $x+1$,
$P(m+1)=(1+x{)}^{m+1}=(1+x{)}^m(1+x)>(1+mx)(1+x),by\left(1\right)$
$=1+(m+1)x+m{x}^2$
$=1+(m+1)x+m{x}^2>1+(m+1)x\because m{x}^2>0$
Above relation shows that P(n) is true for $n=m+1$
Hence P(n) is true universally if $n\ge 2$