I apologize for misunderstanding your question. 
Let me explain how the function f(i) = k f(i-1) + k a can be written as follows:
f(i) = k (f(i-1) + a) = k(k (f(i-2)+ a ) + a) = k (k (k (f(i-3)+ a ) + a) + a) …
Therefore, we have: f(i,t)= (k ^ t) f(i-t) + a k + a k^2 + … + a k^t
The second term of this function is a geometric series minus its first term, where a is the first term and k is the common ratio.
For a geometric series, the sum is: S(n)=b (k ^ n -1) / (k-1), where b is the first term, k is the common ratio, and n is the number of terms.
So the second term of the above function can be written as: a k + a k^2 + … + a k^t = S(t+1) - a=(a (k ^ (t+1) -1) / (k-1))-a
Hence, the final result is: f(i,t) =(k ^ t) f(i-t) + (a (k ^ (t+1) -1) / (k-1))-a.
We can verify this by plugging in t = 1, and we get: f(i,1) = k f(i-1) + a k
This matches the original function.
So the final answer is: f(i,t) =(k ^ t) f(i-t) + (a (k ^ (t+1) -1) / (k-1))-a.
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