The base case is when n=2, so using the given formula we get:
a2=3a1+4a0=24+20=44.
Using the proposed formula we get:
(13/5)(16)+(12/5)(1)=(208+12)/5=220/5=44. The base case holds.
According to the proposed formula, the sum of two consecutive terms of the series causes (12/5)(-1)n to cancel out:
an+an+1=(13/5)4n+(12/5)(-1)n+(13/5)4n+1+(12/5)(-1)n+1,
an+an+1=(13/5)4n(1+4)=13(4n), or an=13(4n-1)-an-1 in general terms.
But, according to the recurrence formula:
an+an+1=3an-1+4an-2+3an+4an-1=3an+7an-1+4an-2=3(3an-1+4an-2)+7an-1+4an-2,
an+an+1=16(an-1+an-2). Applying this for n=2: a2+a3=16(a1+a0)=42(8+5)=42(13)=208. The formula allows us to take consecutive pairs of terms and predict the sum of the next pair.
The series is: 5, 8, 44, 164, ... according to the recursive formula.
If we take the sum of consecutive terms in order we get: 13, 52, 208, ...
This can be written: 13(40), 13(41), 13(42), ...
a1=8=13(40)-5=13(40)-a0, a2=44=13(41)-8=44=13(41)-a1, a3=164=13(42)-44=13(42)-a2, ...
Note that the third sum is 13(42)=208, which concurs with the formula derived from the given recursive formula as applied to sums of consecutive pairs. This could indicate that the second sum is 13(4)=52. (The fifth sum would be 42(42(13))=44(13), and the fourth sum could then be 43(13). We could then speculate the nature of the proposed formula.) We have yet to determine the full proposed formula.
From the proposed formula:
an+1+an=13(4n+1); an-1+an-2=13(4n-1), so an+1+an=16(an-1+an-2) as confirmed earlier using the recursive formula (because 4n+1/4n-1=42=16).
Consider 4 consecutive terms in the series: an, an+1, an+2, an+3.
an+2+an+3=16(an+an+1); an+2=3an+1+4an; an+3=3an+2+4an+1; using the given recursive formula.
Therefore, substituting for an+2, an+3=3(3an+1+4an)+4an+1=13an+1+12an.
Using the proposed formula:
an+3=(13/5)4n+3+(12/5)(-1)n+3; an+1=(13/5)4n+1+(12/5)(-1)n+1; an=(13/5)4n+(12/5)(-1)n.
13an+1=13((13/5)4n+1+(12/5)(-1)n+1)=13(52/5)4n-13(12/5)(-1)n;
12an=12((13/5)4n+(12/5)(-1)n)=12(13/5)4n+12(12/5)(-1)n.
13an+1+12an=(64×13/5)4n-(12/5)(-1)n=(13/5)4n+3+(12/5)(-1)n, which is the expression for an+3 using the proposed formula. Note that (-1)n+3=(-1)n(-1)3=-(-1)n, hence the change of sign.
By induction then, we have the proof of the proposed formula.