8. Use induction (on n) to show that: (a) (2n)! > 2" (n!)?, for n >...
Use mathematical induction to show that when n is an exact power of 2, the solution of the recurrence: { if n 2 2 T(n) for k> 1 if n 2 T(n) 2T(n/2) is T(n) n log
b) Use a mathematical induction to show that: п 2" divides (n + 1) (n + 2) ... (2n – 1) (2n), for n = 0 , 1, 2, ... c) Prove by contradiction: If |x|< ɛ for all ɛ>0, then x = 0.
(2) Prove by induction that for all integers n > 2. Hint: 2n-1-2n2,
Exercise 1.6.4: Prove the following by induction: (a) “k - n(n+1)(2n +1) k= 1 (b) If n > 1, then 13-n is divisible by 3. (c) For n 3, we have n +4 <2". (d) For any positive integer n, one of n, n+2, and 11+ 4 must be divisible by 3. (e) For all n e N, we have 3" > 2n +1. ()/Prove that, for any x > -1 and any n e N, we have (1+x)" 21+1x.
6. Use Mathematical Induction to show that (21 - 1)(2i+1) n for all integers n > 1. 2n +1 (5 marks) i=1
(5) Use induction to show that Ig(n) <n for all n > 1.
2. Prove by induction that Ση.c)-(7+1) for n > 0 and i > 0.
7n Use Mathematical Induction to prove that Σ 2-2n+1-2, for all n e N
9. Prove by mathematical induction: -, i = 1 + 2 + 3+...+ n = n(n+1) for all n > 2.
Prove by mathematical induction that 2-2 KULT = n for all integers n > 2.