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Home / Elementary Number Theory / Prove that the in equality an ≤ (7/4)n holds for every positive integer n.

Prove that the in equality an ≤ (7/4)n holds for every positive integer n.

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

a1 = 1 < (7/4)1 = 7/4

and a2 = 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

ak-1 < (7/4)k -1

and ak-2 < (7/4)k-2

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

ak = ak-1 + ak-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

∴ ak < (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 that an < (7/4)n for all n ≥ 1.

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