Question

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

• A. all $n\in$ N
• B. all $n$ > 1
• C. all $n$ > 1 provided n $\ne$ 0
• D. none of these.

Answer 1

Answer: (C)

For $n=1,(1+x{)}^n+1+x$ and $1+nx-1+x$
$\therefore$ P(1) is not true.
For $n=2,(1+x{)}^n=(1+x{)}^2=1+2x+x^2>1+2x=1+nx$ [$\because x^2\ge$ 0] (for $n$ = 2)
$\therefore$ P ($m$) is true for $n$ = 2
Let P($m$) be true where m is a some +ve integer$\ge$ 2.
$\therefore (1+x{)}^m>1+mx$
$\Rightarrow (1+x{)}^{m+1}>(1+mx)(1+x)$
= $1+(m+1)x+m{x}^2>1+(m+1)x$
$\therefore$ P ($m$ + 1) is true if P ($m$) is true.
$\therefore$ by induction, P($n$) is true for all $n$ > 1 and $x\ne$ 0
[$\because$ for $n=0,(1+x{)}^n$ = 1 and 1 + $nx$ = 1