this post was submitted on 20 Sep 2024
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K•(1+r)^n
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.
The use of for makes sense.
k=0; for (i=0; i<n; i++) k=k+f(i);is the same ask=\sum_{i=0}^{n-1} f(i)and
k=1; for (i=0; i<n; i++) k=k*f(i);is the same ask=\prod_{i=0}^{n-1} f(i)In our case,
f(i)=1+randk=1; for (i=0; i<n; i++) k*(1+r);is the same ask=\prod_{i=0}^{n-1} (1+r) = (1+r)^nAll of that just to say that exponentiation is an iteration of multiplication, the same way that multiplication is an iteration of addition