Question

The inequality $n! > {{2}^{n-1}}$ is true:
A) For all n > 1.
B) For all n > 2
C) For all n$\in $ N
D) None of these.

Answer 1

Hint: First, start by checking the domain of the inequality followed by using the method of mathematical induction to find the answer.

Complete step-by-step answer:
We know that the domain of the function $x!$ is all natural numbers. So, we can say that the domain of the given inequality $n!>{{2}^{n-1}}$ is all natural numbers, but it might be possible that the inequality does not hold for some values of n. So, let us consider that equality holds for all n > k, where k is a natural number.
Now when we take k to be 1 then the smallest possible value of n is 2. But when we put two in the inequality, we find that the inequality is not true. So, we cannot take k to be 1. Again if we take k to be 2 then the smallest possible value of n is 3 and it satisfies the inequality, so we take n > 2 is true.
Now to prove that the inequality is true for all n > 2, we need to use the method of mathematical induction.
So, according to the rule of mathematical induction:
We will check if the relation to be true for the base value, i.e., n=3 and consider the relationship to be true for n=k and then if using this relation, we could prove that the relation holds for n=k+1 then we will say that we have proved the required thing.
Let’s start by taking n=3.
$3! > {{2}^{3-1}}$
$\Rightarrow 6>4$
So, inequality is satisfied. Now letting the inequality be true for n=k, provided n > 2.
$k! > {{2}^{k-1}}$
Now we will multiply both sides of the above inequality by k+1. On doing so, we get
$\left( k+1 \right)k! > \left( k+1 \right){{2}^{k-1}}$
Now we know that k+1 is always greater than 2 for n > 2. So, we can write the inequality as:
$\left( k+1 \right)k!>2\times {{2}^{k-1}}$
$\Rightarrow \left( k+1 \right)!>{{2}^{\left( k+1 \right)-1}}$
So, we have shown that the inequality holds true for n=k+1 as well. Hence, by the rule of mathematical induction, we can say that $n! > {{2}^{n-1}}$ is true for all n > 2.
Therefore, the answer to the above question is option (B).

Note: Don’t forget to check the domain at the starting of such questions as by checking the domain you can eliminate some of the options which might decrease the complexity of the problem. Also be careful while dealing with inequalities as generally students commit mistakes while solving inequalities.