Question

Prove that $(1+x{)}^n\ge (1+nx),forx>-1$ is true for all natural numbers.

Answer 1

Let P (n): $(1+x{)}^n\ge (1+nx),forx>-1$
For n = 1,
L.H.S = $(1+x{)}^1$ = (1 + x)
R.H.S = (1 + 1.x) = (1 + x)
L.H.S $\ge$ R.H.S,
R.H.S,
$\therefore$ P(n) is true for n = 1
Assume P(k) is true
$(1+x{)}^k\ge (1+kx),x>-1$ . . . .(1)
We will prove that P(k+1) is true.
$L.H.S=(1+x{)}^{k+1}$
R.H.S = (1 + (k + 1)x)
$\begin{array}{cc}L.H.S& R.H.S\\ (1+x{)}^{k+1}& (1+(k+1\left)x\right)\\ =(1+x{)}^k(1+x{)}^1& =(1+kx+x)\\ Using\left(1\right):(1+x{)}^k\ge (1+kx)& \\ \ge (1+kx)(1+x)& \\ \ge \left(1\right(1+x)+kx(1+x\left)\right)& \\ \ge (1+x+kx+k{x}^2)& \\ \ge (1+x+kx)+\left(k{x}^2\right)& \\ kisanaturalnumberand{x}^2\ge 0sothatk{x}^2\ge 0.& \\ \ge (1+x+kx)& \end{array}$
L.H.S $\ge$ R.H.S
$\therefore$ P(k + 1) is true whenever P(k) is true.
$\therefore$ By the principle of mathematical induction, P(n) is true for n, where n is a natural number.