Question

Solve the given inequality ${(1 + x)^n} > 1 + nx$ for $n > 2,n \in N$
$x > - 1$ $x \ne 0$.

Answer 1

Hint – Let p(n) be true for n = m. Also, when n = 1, ${(1 + x)^1} = 1 + x$, so in this case inequality does not hold. Use this concept to solve.

Complete step-by-step answer:
We have been given ${(1 + x)^n} > 1 + nx$ for $n > 2,n \in N$.
If $x > - 1$ , so $(x + 1) > 0$ and if we multiply by (x + 1) then the inequality does not change.
Let us assume that P(n) is true for n = m.
Also, if we put n = 1, then-
${(1 + x)^1} = 1 + x$.
This implies that in this case inequality does not hold.
But when n = 2,
${(1 + x)^2} = 1 + 2x + {x^2} > 1 + 2x$
Thus P(n) is true for n = 2.
Now, for n = m –
${(1 + x)^m} > 1 + mx - (1)$
Since, x > -1, multiplying by x + 1, we get-
$P(m + 1) = {(1 + x)^{m + 1}} = {(1 + x)^m}(1 + x) > (1 + mx)(1 + x)$, by equation (1).
$= 1 + (m + 1)x + m{x^2} \\\ = 1 + (m + 1)x + m{x^2} > 1 + (m + 1)x,\\{ \because m{x^2} > 0\\} \\\$
Above relation shows that P(n) is true for n = m + 1.
Hence, P(n) is true universally if $n \geqslant 2$.

Note – Whenever such types of questions occur, then always put different values of n to see at what values of n does the inequality satisfy, as mentioned in the solution that for n = 1, the inequality does not hold but for n = 2 it holds. And then, we found P(m+1), and we see that P(n) is true universally if $n \geqslant 2$.