By Cruz F. R., Mateus G. R., Smith J. M.

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**Example text**

4. Yes, I realize that you can solve this without induction by applying some elementary algebra and the identities of Problems 1 and 2. What I want you to do is practice your induction by proving this identity from ﬁrst principles. 5. See the comment to Problem 4. 6. See the comment to Problem 4. 9. There is an elegant noninductive proof of this that you should have met in high school. Let S = ni=0 ai . Then, n+1 n ai − aS − S = i=1 ai = an+1 − 1. i=0 Hence, S = (an+1 − 1)/(a − 1), as required.

1) 42 Chap. 4. 1) is true by induction on i. It is trivially true for i = 1. Now suppose that i > 1 and that i−2 T (n) = 3i−1 T (n − i + 1) + 2 3j . j=0 Then, i−2 T (n) = 3i−1 T (n − i + 1) + 2 3j j=0 i−2 = 3i−1 (3T (n − i) + 2) + 2 3j j=0 i−1 = 3i T (n − i) + 2 3j , j=0 as required. 1), we can continue with solving the recurrence. Suppose we take i = n − 1. 1), n−2 n−1 T (n) = 3 3j T (1) + 2 j=0 = 3n−1 + 3n−1 − 1 = 2 · 3n−1 − 1. (by Problem 9) 208. Suppose T (1) = 1, and for all n ≥ 2 a power of 2, T (n) = 2T (n/2) + 6n − 1.

242. Suppose T (1) = 1, and for all n ≥ 2, T (n) = T ( n/2 ) + T ( n/2 ) + n − 1. Find an exact solution for T (n). 243. Solve the following recurrence relation exactly. T (1) = 1 and for all n ≥ 2, T (n) = T ( n/2 ) + 1. 244. Solve the following recurrence relation exactly. T (1) = 1 and for all n ≥ 2, T (n) = T ( n/2 ) + 1. 245. Solve the following recurrence relation exactly. T (1) = 1, and for n ≥ 2, T (n) = 2T ( n/2 ) + 6n − 1. 246. Solve the following recurrence relation exactly. T (1) = 2, and for all n ≥ 2, T (n) = 4T ( n/3 ) + 3n − 5.