University:
Whitman College
Course:
MATH-126 | Calculus II
Academic year:

2024
Views:

487

Pages:

1
Author:

banenetta0l

Exercises in Proof by Induction Here’s a summary of what we mean by a “proof by induction”: The Induction Principle: Let P (n) be a statement which depends on n = 1, 2, 3, · · ·. Then P (n) is true for all n if: • P (1) is true (the base case). • Prove that P (k) is true implies that P (k + 1) is true. This is sometimes broken into two steps, but they go together: Assume that P (k) is true, then show that with this assumption, P (k + 1) must be true. Exercises 1. Prove each using induction: (a) n X i= i=1 n X 2 n X i−1 n(n + 1) 2 (e) n(n + 1)(2n + 1) (b) i = 6 i=1 (c) i=1 n X 2 (f) (g) n =2 −1 1 1 1 n + +···+ = 1·2 2·3 n(n + 1) n+1 n X 1 n = n+1 i=1 i(i + 1) n X (2i − 1) = n2 i=1 (h) n! > 2n for n ≥ 4. n2 (n + 1)2 (d) i = 4 i=1 3 (i) 2n+1 > n2 for all positive integers. 2. This exercise refers to the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, · · · The sequence is defined recursively by f1 = 1, f2 = 1, then fn+1 = fn + fn−1 for each n > 2. As before, prove each of the following using induction. You might investigate each with several examples before you start. (a) f1 + f2 + · · · + fn = fn+2 − 1 (b) f12 + f22 + · · · + fn2 = fn fn+1 (c) f1 + f3 + f5 + · · · + f2n−1 = f2n 1

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