Weird Recursion-1

Algebra Level 3

A recurrence relation is given by b 0 = 2 b_0=2 and b n + 1 b n − 4 b n + 4 = 0 b_{ n+1 }b_{ n }-4b_{ n }+4=0 for n > 0 n>0 . Find b n b_n .


The answer is 2.

Ahmed Arup Shihab by Ahmed Arup Shihab , 28, Bangladesh

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1 solution

Tom Engelsman
Feb 12, 2017

Let us rewrite the above recurrence relation as:

b n + 1 b n = 4 b n − 4 ⇒ b n + 1 = 4 − 4 b n . b_{n+1}b_{n} = 4b_{n} - 4 \Rightarrow b_{n+1} = 4 - \frac{4}{b_{n}}.

Starting with b 0 = 2 b_0 = 2 , we now obtain:

b 1 = 4 − 4 b 0 = 2 ; b_1 = 4 - \frac{4}{b_0} = 2;

b 2 = 4 − 4 b 1 = 2 ; b_2 = 4 - \frac{4}{b_1} = 2;

b 3 = 4 − 4 b 2 = 2 ; b_3 = 4 - \frac{4}{b_2} = 2;

and so forth. Thus b n = 2 b_n = 2 for all n ∈ N . n \in \mathbb{N}.

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