4 ms·
This is the correct way to prove the summation. Although, I would point out that for r > 1, it's probably better to phrase it as ( r^(n+1) - 1)/(r-1). However,
by Locke1689 17y ago
This is the correct way to prove the summation. Although, I would point out that for r > 1, it's probably better to phrase it as ( r^(n+1) - 1)/(r-1). However, it's important to note that this is only true from a series starting from zero. For a series starting from m it is:
(r^(n+1)-r^m)/(r-1)
Which is trivially deducible from your method or the parent's method. For some reason I tend to dislike geometric proofs. This is probably because I find it very easy to make proof fallacies visually, more so than symbolically. However, this geometric proof is also less useful than the symbolic because it is much harder to use it to prove the summation at values smaller than infinity, whereas the symbolic works just as well for n.