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this post was submitted on 20 Sep 2024
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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+r
andk=1; for (i=0; i<n; i++) k*(1+r);
is the same ask=\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