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464 Chap. 14 Analysis TechniquesDividing by both n and f(n − 1) happen to give us useful patterns tof(n)work with.n= n+1 f(n)2, andf(n−1)= n+1n−1. Of course, lots of otherguesses for function g do not work. For example, f(n) − n = f(n −1). Knowing that f(n) = f(n − 1) + n is not useful for determining theclosed form solution to this summation. Or consider f(n) − f(n − 1) = n.Again, knowing that f(n) = f(n − 1) + n is not useful. Finding the rightcombination of equations can be like finding a needle in a haystack.In our first example, we can see directly what the closed-form solutionshould be. Since f(n)n = n+12, obviously f(n) = n(n + 1)/2.Dividing f(n) by f(n − 1) does not give so obvious a result, but itprovides another useful illustration.f(n)f(n − 1)= n + 1n − 1f(n)(n − 1) = (n + 1)f(n − 1)f(n)(n − 1) = (n + 1)(f(n) − n)nf(n) − f(n) = nf(n) + f(n) − n 2 − n2f(n) = n 2 + n = n(n + 1)f(n) =n(n + 1)2Once again, we still do not have a proof that f(n) = n(n+1)/2. Why?Because we did not prove that f(n)/n = (n + 1)/2 nor that f(n)/f(n −1) = (n + 1)(n − 1). We merely hypothesized patterns from looking at afew terms. Fortunately, it is easy to check our hypothesis with induction.Example 14.2 Solve the summationn∑1/2 i .i=1We will begin by writing out a table listing the first few values of the summation,to see if we can detect a pattern.n 1 2 3 4 5 61f(n)211 − f(n)234147818151611631321326364164