Using the principle of mathematical induction, prove each of the following

Solution:

To Prove:

$3^{n} \geq 2^{n}$

Let us prove this question by principle of mathematical induction (PMI) for all natural numbers

Let $\mathrm{P}(\mathrm{n}): 3^{n} \geq 2^{n}$

For $n=1 P(n)$ is true since $3^{n} \geq 2^{n} i \times e \times 3 \geq 2$, which is true

Assume P(k) is true for some positive integer k , ie

$=3^{k} \geq 2^{k} \ldots(1)$

We will now prove that P(k + 1) is true whenever P( k ) is true

Consider,

$=3^{(k+1)}$

$\therefore 3^{(k+1)}=3^{k} \times 3>2^{k} \times 3$ [Using 1]

$=3^{k} \times 3>2^{k} \times 2 \times \frac{3}{2}$ [Multiplying and dividing by 2 on RHS]

$=3^{k+1}>2^{k+1} \times \frac{3}{2}$

Now, $2^{k+1} \times \frac{3}{2}>2^{k+1}$

$\therefore 3^{k+1}>2^{k+1}$

Therefore, P (k + 1) is true whenever P(k) is true

By the principle of mathematical induction, P(n) is true for all natural numbers, ie, N.

Hence proved.