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Sec. 14.2 Recurrence Relations 473Note that= a m T(1) + a m−1 c(n/b m−1 ) k + · · · + ac(n/b) k + cn k= a m c + a m−1 c(n/b m−1 ) k + · · · + ac(n/b) k + cn km∑= c a m−i b iki=0∑m= ca m (b k /a) i .i=0a m = a log b n = n log b a . (14.1)The summation is a geometric series whose sum depends on the ratio r = b k /a.There are three cases.1. r < 1. From Equation 2.4,Thus,m∑r i < 1/(1 − r), a constant.i=0T(n) = Θ(a m ) = Θ(n log ba ).2. r = 1. Because r = b k /a, we know that a = b k . From the definitionof logarithms it follows immediately that k = log b a. We also note fromEquation 14.1 that m = log b n. Thus,m∑r = m + 1 = log b n + 1.i=0Because a m = n log b a = n k , we have3. r > 1. From Equation 2.5,T(n) = Θ(n log b a log n) = Θ(n k log n).Thus,m∑i=0r = rm+1 − 1r − 1= Θ(r m ).T(n) = Θ(a m r m ) = Θ(a m (b k /a) m ) = Θ(b km ) = Θ(n k ).We can summarize the above derivation as the following theorem, sometimesreferred to as the Master Theorem.