NOTE
1,1,2,3,5,.......... is a fibonacci series.
Name the above fibonacci series as fibonacci 1.
Let us assume the nth term of fibonacci 1 be fn
then fn = 1/√5 ( ( (1+√5)/2 )n - ( ( 1-√5)/2 )n ) (Consider this as a fact)
Now, come to our problem
Let us assume T(n) = akn bmn ........................................................... (1)
then T(n+1) = akn+1 bmn+1...............................................(2)
Now, it is given in the question that T(n+2) = T(n) * T(n+1)
=> T(n+2) = akn bmn * akn+1 bmn+1 from equations 1 & 2
=> T (n+2) = a(kn + kn+1) b(mn + mn+1)
So, we are seeing that the power of 'a' in T(n+2) = summation of the powers of 'a' in T(n) and T(n+1)
Now, T (n+2) = akn+2 bmn+2
So, we get kn+2 = kn +kn+1
Thus the powers of 'a' forms a fibonacci series.
Similarly we can say it for the powers of 'b'
here mn+2 = mn +mn+1
Now, T(1) = a so, k1 = 1 and m1 = 0
Again T(2) = b so, k2 = 0 and m2 = 1
Thus the powers of 'a' forms the fibonacci series 1,0,1,1,2,......
name the above fibonacci series as fibonacci 2
If we look at the series fibonacci 1 and fibonacci 2 carefully we get that the nth term of fibonacci 2 is equal to (n-2)th term of fibonacci 1 for n>=3.
So, the terms of fibonacci 2 are
k1 = 1 k2 = 0 and kn = fn-2 for n>=3.
Similarly we get that the powers of 'b' forms the fibonacci series 0,1,1,2,.... name this series as Fibonacci 3
We get that nth term of fibonacci 3 is equal to (n-1)th term of Fibonacci 1
so, the terms of Fibonacci 3 are
m1 = 0, m2=1 , mn = fn-1 for n>=3
So, our required solution T(n) = akn bmn = afn-2bfn-1 for n>=3.