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Sec. 2.4 Summations and Recurrences 33We can expand the recurrence as many steps as we like, but the goal isto detect some pattern that will permit us to rewrite the recurrence in termsof a summation. In this example, we might notice that(T(n − 2) + 1) + 1 = T(n − 2) + 2and if we expand the recurrence again, we getT(n) = T(n − 2) + 2 = T(n − 3) + 1 + 2 = T(n − 3) + 3which generalizes to the pattern T(n) = T(n − i) + i. We might concludethatT(n) = T(n − (n − 1)) + (n − 1)= T(1) + n − 1= n − 1.Because we have merely guessed at a pattern and not actually provedthat this is the correct closed form solution, we should use an inductionproof to complete the process (see Example 2.13).Example 2.9 A slightly more complicated recurrence isT(n) = T(n − 1) + n; T (1) = 1.Expanding this recurrence a few steps, we getT(n) = T(n − 1) + n= T(n − 2) + (n − 1) + n= T(n − 3) + (n − 2) + (n − 1) + n.We should then observe that this recurrence appears to have a pattern thatleads toT(n) = T(n − (n − 1)) + (n − (n − 2)) + · · · + (n − 1) + n= 1 + 2 + · · · + (n − 1) + n.This is equivalent to the summation ∑ nclosed-form solution.i=1i, for which we already know theTechniques to find closed-form solutions for recurrence relations are discussedin Section 14.2. Prior to Chapter 14, recurrence relations are used infrequently inthis book, and the corresponding closed-form solution and an explanation for howit was derived will be supplied at the time of use.