Determine the general term of sequences (un) defined by:4un+2−4un+1+un=0,u0=2,u1=3

Question

4un+2−4un+1+un=0,u0=2,u1=3

Answer 1

4u_n+2-4u_n+1+u_n=0, u₀=2, u₁=3.

Let's find some terms.

4u₂-4u₁+u₀=0, u₂=(4u₁-u₀)/4=(12-2)/4=5/2.

4u₃-4u₂+u₁=0, u₃=(4u₂-u₁)/4=(10-3)/4=7/4.

4u₄-4u₃+u₂=0, u₄=(4u₃-u₂)/4=(7-5/2)/4=9/8.

So, so far we have: 2, 3, 5/2, 7/4, 9/8, ... which can be written:

1/½, 3/1, 5/2, 7/4, 9/8, ... 

A pattern suggest itself: The numerators form a sequence of natural odd numbers, while the denominators form a sequence of powers of 2. So:

u_n=(2n+1)/2^n-1. On this assumption, u₅=11/16.

4u₅-4u₄+u₃=0, u₅=(4u₄-u₃)/4=(9/2-7/4)/4=11/16. This confirms u_n=(2n+1)/2^n-1.