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Sec. 14.1 Summation Techniques 463which in turn yield c 1 = 1/2 and c 2 = 1/2. Thus, if the closed-form solution forthe summation is a polynomial, it can only bewhich is more commonly written1/2n 2 + 1/2n + 0n(n + 1).2At this point, we still must do the “test” part of the guess-and-test approach. Wecan use an induction proof to verify whether our candidate closed-form solution iscorrect. In this case it is indeed correct, as shown by Example 2.11. The inductionproof is necessary because our initial assumption that the solution is a simplepolynomial could be wrong. For example, it might have been that the true solutionincludes a logarithmic term, such as c 1 n 2 + c 2 n log n. The process shown here isessentially fitting a curve to a fixed number of points. Because there is always ann-degree polynomial that fits n + 1 points, we have not done enough work to besure that we to know the true equation without the induction proof.Guess-and-test is useful whenever the solution is a polynomial expression. Inparticular, similar reasoning can be used to solve for ∑ n∑ i=1 i2 , or more generallyni=1 ic for c any positive integer. Why is this not a universal approach to solvingsummations? Because many summations do not have a polynomial as their closedform solution.A more general approach is based on the subtract-and-guess or divide-andguessstrategies. One form of subtract-and-guess is known as the shifting method.The shifting method subtracts the summation from a variation on the summation.The variation selected for the subtraction should be one that makes most of theterms cancel out. To solve sum f, we pick a known function g and find a pattern interms of f(n) − g(n) or f(n)/g(n).Example 14.1 Find the closed form solution for ∑ ni=1i using the divideand-guessapproach. We will try two example functions to illustrate thedivide-and-guess method: dividing by n and dividing by f(n − 1). Ourgoal is to find patterns that we can use to guess a closed-form expression asour candidate for testing with an induction proof. To aid us in finding suchpatterns, we can construct a table showing the first few numbers of eachfunction, and the result of dividing one by the other, as follows.n 1 2 3 4 5 6 7 8 9 10f(n) 1 3 6 10 15 21 28 36 46 57n 1 2 3 4 5 6 7 8 9 10f(n)/n 2/2 3/2 4/2 5/2 6/2 7/2 8/2 9/2 10/2 11/2f(n−1) 0 1 3 6 10 15 21 28 36 46f(n)/f(n−1) 3/1 4/2 5/3 6/4 7/5 8/6 9/7 10/8 11/9