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 $\neq$ 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 \geq $ 0] (for $n$ = 2)
$\therefore $ P ($m$) is true for $n$ = 2
Let P($m$) be true where m is a some +ve integer$\geq$ 2.
$\therefore \, (1 + x)^m > 1 + mx$
$\Rightarrow \, (1 + x)^{m+1} > (1 + mx) (1 + x) $
= $1 + (m + 1) x + mx^2 > 1 + (m + 1)x$
$\therefore $ P ($m$ + 1) is true if P ($m$) is true.
$\therefore $ by induction, P($n$) is true for all $n$ > 1 and $x \, \neq $ 0
[$\because$ for $n = 0, (1 + x)^n$ = 1 and 1 + $nx$ = 1