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1 + 2 + … + n = n(n+1)/2. That is worth remembering; it comes up a lot in analysis of algorithms.
Obviously, n(n+1)/2 can be computed using only a few steps, so repetition is not really required to compute 1 + 2 + … + n. But for the purposes of these notes, computing 1 + 2 + … + n is a simple problem that serves to illustrate repetition.
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