By Barutello V., Terracini S.

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**Example text**

The claim is certainly true for n = 0, in which case both sides of the equation are zero. Suppose that n ≥ 0 and ni=1 i = n(n + 1)/2. We are required to prove that n+1 i=1 i = (n + 1)(n + 2)/2. Now, n+1 n i = i=1 i + (n + 1) i=1 = n(n + 1)/2 + (n + 1) = (n + 1)(n + 2)/2, (by the induction hypothesis) as required. 8. n We are required to prove that for all n ≥ 1, i=1 1/2i = 1 − 1/2n . The claim is true for n = 1, since in this case both sides of the equation are equal to 1/2. Now suppose that n ≥ 1, and ni=1 1/2i = 1 − 1/2n .

Suppose f : IR → IR is continuous and monotonically increasing, and has the property that if f (x) ∈ ZZ, then x ∈ ZZ. Then, f ( x ) = f (x) and f ( x ) = f (x) . 248. 76 log n/ log log n, but this may not help you much. 7 SOLUTIONS 202. Suppose T (1) = 1, and for all n ≥ 2, T (n) = 3T (n − 1) + 2. If n is large enough, then by repeated substitution, T (n) = = = = = 3T (n − 1) + 2 (after one substitution) 3(3T (n − 2) + 2) + 2 9T (n − 2) + 2 · 3 + 2 (after two substitutions) 9(3T (n − 3) + 2) + 2 · 3 + 2 27T (n − 3) + 2 · 9 + 2 · 3 + 2 (after three substitutions).

Function max(x, y) comment Return largest of x and y. if x ≥ y then return(x) else return(y) 54 275. Chap. 5. n] is correct. 1. 2. 3 276. Fibonacci Numbers Prove that the following algorithm for computing Fibonacci numbers is correct. function ﬁb(n) comment Return Fn , the nth Fibonacci number. if n ≤ 1 then return(n) else return(ﬁb(n − 1)+ﬁb(n − 2)) 1. 2. 277. n]. if n ≤ 1 then return(A[1]) else return(sum(n − 1) + A[n]) Prove that the following algorithm for computing Fibonacci numbers is correct.