4 points
1 point
I would just rebuild something in my head like this every time.
While i < n; k=k+(k*r); i++;
You’d think I could remember k(1+r)^n but when you posted, it looked as alien as it felt decades ago.
5 points
The use of for makes sense.
k=0; for (i=0; i<n; i++) k=k+f(i);
is the same as
k=\sum_{i=0}^{n-1} f(i)
and
k=1; for (i=0; i<n; i++) k=k*f(i);
is the same as
k=\prod_{i=0}^{n-1} f(i)
In our case, f(i)=1+r and
k=1; for (i=0; i<n; i++) k*(1+r);
is the same as
k=\prod_{i=0}^{n-1} (1+r) = (1+r)^n
All of that just to say that exponentiation is an iteration of multiplication, the same way that multiplication is an iteration of addition
