Solution: First of all for n = 1 and 2, we have

a_{1} = 1 < (7/4)^{1} = 7/4

and a_{2} = 3 < (7/4)^{2} = 49/16

Where the inequality in question holds in this two cases. This provides a basis for the induction. For the induction step, pick an integer k ≥ 3 and assume that the inequality is valid for n =1, 2, 3, …, k -1 .

Then in particular

a_{k-1} < (7/4)^{k -1}

and a_{k-2} < (7/4)^{k-2}

By the way in which the Lucas sequence is formed, it follows that

a_{k} = a_{k-1} + a_{k-2} < (7/4)^{k-1} + (7/4)^{k-2}

= (7/4)^{k-2} (7/4 +1)

= (7/4)^{k-2} (11/4)

< (7/4)^{k-2} (7/4)^{2}

= (7/4)^{k}

∴ a_{k} < (7/4)^{k}

### Since the inequality is true for n = k. Whenever it is true for the integers

1, 2, …,k -1. *We conclude by the seemed induction principle tha*t a_{n} < (7/4)^{n} for all n ≥ 1.