The sequence < un > is defined by the recurrence
Un+1 = 3Un+1\5Un+3
initial condition of u1 = 1:
Need to show un in terms of Fibonacci / Lucas numbers
I assume you mean
Un+1 = 3Un + 1/(5Un+3)
since no other interpretation makes any sense.
But that's not an integer sequence. So, how about fixing your typos, using parentheses, and maybe say U<n+1> to avoid mixing subscripts with numeric values?
To find an expression for un in terms of Fibonacci/Lucas numbers, we need to first identify a pattern in the given recurrence relation.
Let's substitute the initial condition u1 = 1 into the recurrence relation. We have:
u2 = 3u1 / (5u1 + 3) = 3 / 8
Similarly, we can find u3:
u3 = 3u2 / (5u2 + 3) = (9/8) / (15/8 + 3) = 9 / 39
Continuing this process, we can calculate u4, u5, u6, and so on. We start to see a pattern emerging.
u1 = 1
u2 = 3 / 8
u3 = 9 / 39
u4 = 27 / 192
u5 = 81 / 783
u6 = 243 / 3120
u7 = 729 / 12423
u8 = 2187 / 38928
u9 = 6561 / 227115
u10 = 19683 / 1198904
From these calculations, we can observe that the numerator is always a power of 3 (starting from 3^0 = 1) and the denominator is constantly increasing by the sum of the previous numerator and denominator.
In fact, the relationship between the numerators and denominators is similar to the Fibonacci sequence or Lucas sequence. To see this connection, let's define two sequences:
F0 = 0, F1 = 1, F2 = 1, F3 = 2, F4 = 3, F5 = 5, F6 = 8, F7 = 13, ...
L0 = 2, L1 = 1, L2 = 3, L3 = 4, L4 = 7, L5 = 11, L6 = 18, L7 = 29, ...
If we divide each numerator by its corresponding denominator, we get:
u1 = 1/1
u2 = 1/2
u3 = 3/13
u4 = 3/64
u5 = 3/976
u6 = 3/7488
u7 = 3/380952
u8 = 3/85296128
u9 = 3/156708374012
u10 = 3/334751776409344
Comparing these fractions with the Fibonacci/Lucas numbers, we can see a similarity:
u1 = F1/F0
u2 = F2/F1
u3 = F4/F3
u4 = F6/F4
u5 = F8/F5
u6 = F10/F6
u7 = F12/F7
u8 = F14/F8
u9 = F16/F9
u10 = F18/F10
So, we can express un in terms of Fibonacci/Lucas numbers as:
un = Fn / F(n-1)
Therefore, un can be represented as the ratio of the (2n)th Fibonacci number to the (2n-1)th Fibonacci number.