1 2 3 ... n=n(n 1)/2 proof 616020-1+2+3+...+n=n(n+1)/2 proof

Solved How Can You Show 3 6 9 3n 3n N 1 2 Is True For All Natural Numbers Using Proof By Mathematical Induction Course Hero

A proof of the binomial theorem If n is a natural number, let n! Two, we assume that it is true for n = k and prove that if it is true for n = k, then it is also true for n = k 1 First Step − Now for 1 1 ⋅ 2 1 2 ⋅ 3 1 n(n 1) = n n 1, we know for n = 1, we have 1 1 ⋅ 2 = 1 2 and for n = 2, we have 1 1 ⋅ 2 1 2 ⋅ 3 = 1 2 1 6 = 2 3 = 2 2 1

1+2+3+...+n=n(n+1)/2 proof

1+2+3+...+n=n(n+1)/2 proof-Example(countingpairs) Weprovethatk n k =n n−1 k−1 TheLHScountsthenumberofpairs(x,S),whereS isaksubsetof{1,2,,n} and x ∈ SThere are n k choices for k, andThe sum of the first n squares, 1 2 2 2 n 2 = n(n1)(2n1)/6 For example, 1 2 2 2 10 2 =10×11×21/6=385 This result is usually proved by a method known as mathematical induction, and whereas it is a useful method for showing that a formula is true, it does not offer any insight into where the formula comes from

this involves the following steps ∙ prove true for some value, say n = 1 ∙ assume the result is true for n = k ∙ prove true for n = k 1 n = 1 → LH S = 12 = 1 and RHS = 1 6 (1 1)(2 1) = 1 ⇒result is true for n = 1 assume result is true for3 Prove that (3n (3n — is always a multiple of 12, for all positive integer values of n (Total for question 3 is 2 marks) 4 n is an integer Prove algebraically that the sum of n(n 1) and n 1 is always a square number (Total for question 4 is 2 marks) For induction, you have to prove the base case Then you assume your induction hypothesis, which in this case is 2 n >= n 2 After that you want to prove that it is true for n 1, ie that 2 n1 >= (n1) 2 You will use the induction hypothesis

= 1 ·2 = 2, 3!100 100 positive integers, Gauss quickly used a formula to calculate the sum of 5050 5050 5050 The formulas for the first few values of a a a are as follows ∑ k = 1 n k = n ( n 1) 2 ∑ k = 1 n k 2 = n ( n 1) ( 2 n 1) 6 ∑ k = 1 n k 3 = n 2 ( n 1) 2 4And so the domain of this function is really all positive integers N has to be a positive integer And so we can try this out with a few things, we can take S of 3, this is going to be equal to 1 plus 2 plus 3, which is equal to 6 We could take S of 4, which is going to be 1 plus 2 plus 3 plus 4, which is going to be equal to 10

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Induction Examples Question 7 Consider the famous Fibonacci sequence fxng1 n=1, de ned by the relations x1 = 1, x2 = 1, and xn = xn 1 xn 2 for n 3 (a) Compute x (b) Use an extended Principle of Mathematical Induction in order to show that for n 1, xn = 1 Basic Step When n = 10, the inequality 2^n > n^3 is true as 2^10 = 1024 > 1000 Inductive Step Assume n = k is true, where k ≥ 10,

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