Question

Given ${(1 + x)^n} \geqslant (1 + nx).$ for all natural numbers n, where $x > - n$. What will be the value of n.

Answer 1

Hint: Natural numbers are those used for counting and ordering. In common mathematical terminology words which are used for counting are called “cardinal numbers” and used for ordering is called “ordinal number”. It doesn’t include zero (0). Natural numbers are positive integers or non-negative integers.

Complete step by step solution:
Given equation,
${(1 + x)^n} \geqslant (1 + nx).$
$P(n):{(1 + x)^n} \geqslant (1 + nx)\,when\,n \in N\,and\,x > - n$
Let n=1
Then
$P(1):(1 + x) \geqslant (1 + x)\,$
which is true
$\therefore P(1)\,$is true.
Let us consider $P(m)\,$is true
$ \Rightarrow {(1 + x)^m} \geqslant (1 + mx)$ …(1)
$\therefore x > - n$
Then
$ \Rightarrow x + n > 0$
Now if we take n=1.
$ \Rightarrow x + 1 > 0$
Multiplying both side by (x+1) in equation 1
$ \Rightarrow {(1 + x)^{m + 1}} \geqslant (1 + mx)(1 + x)$
$ \Rightarrow {(1 + x)^{m + 1}} \geqslant 1 + (m + 1)x + m{x^2}$
$ \Rightarrow {(1 + x)^{m + 1}} \geqslant 1 + {(m + 1)_x}$ …(2)
$\therefore m{x^2} \geqslant 0$
$\therefore P(n)$ is true for all values of $n \in N$ and \[x >-1\].
Therefore, the value of $n=1$.

Note:
The number system or numeral system is the system of naming. There are various types of number systems. Natural numbers are counting numbers excluding 0. Whole numbers are all natural numbers including 0. Integers include all positive and negative numbers. Rational number which can be expressed in the form of $\dfrac{p}{q}$ where q≠0. Real numbers are those numbers which can be written in decimals also.