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submitted 3 months ago by [email protected] to c/[email protected]

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[–] [email protected] 4 points 3 months ago (1 children)

K•(1+r)^n

[–] [email protected] 1 points 3 months ago (1 children)

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.

[–] [email protected] 5 points 3 months ago

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