Question

For every positive integral value of n, ${3^n} > {n^3}$ when:
A. $n > 2$
B. $n \geqslant 3$
C. $n \geqslant 4$
D. $n < 4$

Answer 1

Hint:A proof by induction consists of two cases.
o The first, the base case (or basis), proves the statement for n = 0 without assuming any
knowledge of other cases.

o The second case, the induction step, proves that if the statement holds for any given
case $n = k$, then it must also hold for the next case $n = k + 1$.
These two steps establish that the statement holds for every natural number n. The base case does not necessarily begin with $n = 0$, but often with $n = 1$, and possibly with any fixed natural number $n = N$, establishing the truth of the statement for all the natural numbers $n \geqslant N$.

Complete step by step Solution:
Considering the positive integer values of n, we observe the following:
$n = 1$, ${3^1} = 3 > {1^3} = 1$
$n = 2$, ${3^2} = 9 > {2^3} = 8$
$n = 3$, ${3^3} = 27 = {3^3}$
$n = 4$, ${3^4} = 81 > {4^3} = 64$
$n = 5$, ${3^5} = 243 > {5^3} = 125$
And so on.
Since, ${3^n} = {n^3}$ for $n = 3$, there is a good chance, that ${3^n} > {n^3}$ for $n \geqslant 4$.

To, verify, let us use the principle of mathematical induction.
Let's say that the statement is $P(n):\;{3^n} > {n^3}$.
Basis: $P(4):\;{3^4} > {4^3}$ is true.
Induction: Let's say that $P(k):\;{3^k} > {k^3}$ is true for some $k \geqslant 4$.
We have, $P(k + 1):\;{3^{k + 1}} = {3.3^k} > 3.{k^3} = {(k + 1)^3} + (2k - 5)({k^2} + k + 1) + 4 > {(k + 1)^3}$,
for $k > \dfrac{5}{2}$.

Hence, $P(n):\;{3^n} > {n^3}$ is true for all positive integers $n \geqslant 4$.

The correct answer is C. $n \geqslant 4$.

Note:If the inequality reverses between two numbers, then it must also be equal at some point between them.
Since natural numbers are involved, quickly verify the inequality using mathematical induction to see if it is true for all positive integers in a specific open interval or not.