Question

Prove by induction the inequality $(1+x{)}^n\ge 1+nx,$ whenever x is positive and n is a positive integer.

Answer 1

Let p(n) be the statement given by

$P\left(n\right):(2+x)n\ge 1+nx$

For $n=1,P\left(1\right):(1+x{)}^1\ge 1+x$

$\because 1+x=1+x$

$\therefore P\left(1\right)$is true.

Assume that , P(k) is true for some natural number k.

Then $P\left(k\right):(1+x{)}^k\ge 1+kx$

We shall now show that . P(k+1)is true, whenever P(k) is true.

i.e. $P(k+1{)}^{k+1}\ge 1+(k+1)x$

Now consider P(K) is true.

$\therefore (1+x{)}^k\ge 1+kx$

$\Rightarrow (2+x{)}^k(1+x)\ge (1+k\left)\right(1+x)$

[multiplying both sides by (1+x)]

$\Rightarrow (1+x{)}^{k+1}\ge 1+(k+1)x+k{x}^2$

$\Rightarrow (1+x{)}^{k+1}\ge 1+(k+1\left)\right(x+k{x}^2\ge 1+(k+1)x$

$[\because k{x}^2>0]$

$\Rightarrow (1+x{)}^{k+1}\ge 1+(k+1)x$

$\Rightarrow P(k+1)$is true.

Hence, by the principle of mathematical induction P(n) is true,$\forall n\in N.$