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## 1+2+3+…+n = n(n+1)/2

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**1+2+3+…+n = n(n+1)/2**We want to prove the above statement by mathematical Induction for all natural numbers (n=1,2,3,…) Slide 1 Next Slide**Step #1: Prove for n=1**1+2+3+…+n = n(n+1)/2 We need to show that for n=1 the statement is true. Previous Slide Slide 2 Next Slide**Step #1: Prove for n=1**1 | 1(1+1)/2 1+2+3+…+n = n(n+1)/2 Left hand side: Since when n=1, the first and last terms are the same, there is only 1 term on the left hand side. Right hand side: Plug in 1 for n. Previous Slide Slide 3 Next Slide**Step #1: Prove for n=1**1 | 1(1+1)/2 | 2/2 = 1 1+2+3+…+n = n(n+1)/2 Simplify the right hand side to show that both sides are equal. Previous Slide Slide 4 Next Slide**Step #1: Prove for n=1**1 | 1(1+1)/2 | 2/2 = 1 Step #2: Assume True for n=k 1+2+3+…+n = n(n+1)/2 Since we have only proven the statement is true for n=1, we now are going to assume that it is true for an arbitrary natural number k. The word “Assume” is required in the proof to say that we are temporarily assuming this to be true even though we have not proven it yet. Previous Slide Slide 5 Next Slide**Step #1: Prove for n=1**1 | 1(1+1)/2 | 2/2 = 1 Step #2: Assume True for n=k 1+2+3+…+k = k(k+1)/2 1+2+3+…+n = n(n+1)/2 As far as work for step #2, we just substitute k for n in the statement. Previous Slide Slide 6 Next Slide**Step #1: Prove for n=1**1 | 1(1+1)/2 | 2/2 = 1 Step #2: Assume True for n=k 1+2+3+…+k = k(k+1)/2 Step #3: Using step #2, prove for n=k+1 1+2+3+…+n = n(n+1)/2 We now make use of our assumption in step #2 to prove for the next value of n, k+1. Previous Slide Slide 7 Next Slide**Step #1: Prove for n=1**1 | 1(1+1)/2 | 2/2 = 1 Step #2: Assume True for n=k 1+2+3+…+k = k(k+1)/2 Step #3: Using step #2, prove for n=k+1 1+2+3+…+(n-1)+n | n(n+1)/2 1+2+3+…+n = n(n+1)/2 Before we substitute k+1 for n, it’s a good idea to rewrite the statement showing the second to last term on the left hand side. Since the terms keep increasing by 1, and the last term is n, the second to last term is (n-1). I put this in parenthesis to show it as one term. Previous Slide Slide 8 Next Slide**Step #1: Prove for n=1**1 | 1(1+1)/2 | 2/2 = 1 Step #2: Assume True for n=k 1+2+3+…+k = k(k+1)/2 Step #3: Using step #2, prove for n=k+1 1+2+3+…+(n-1)+n | n(n+1)/2 1+2+3+…+(k+1-1)+(k+1) | (k+1)(k+1+1)/2 1+2+3+…+n = n(n+1)/2 Now we’ll substitute k+1 in for n. Previous Slide Slide 9 Next Slide**Step #1: Prove for n=1**1 | 1(1+1)/2 | 2/2 = 1 Step #2: Assume True for n=k 1+2+3+…+k = k(k+1)/2 Step #3: Using step #2, prove for n=k+1 1+2+3+…+(n-1)+n | n(n+1)/2 1+2+3+…+(k+1-1)+(k+1) | (k+1)(k+1+1)/2 1+2+3+…+k+(k+1) | (k+1)(k+2)/2 1+2+3+…+n = n(n+1)/2 Simplify each side. Previous Slide Slide 10 Next Slide**Step #1: Prove for n=1**1 | 1(1+1)/2 | 2/2 = 1 Step #2: Assume True for n=k 1+2+3+…+k = k(k+1)/2 Step #3: Using step #2, prove for n=k+1 1+2+3+…+(n-1)+n | n(n+1)/2 1+2+3+…+(k+1-1)+(k+1) | (k+1)(k+1+1)/2 1+2+3+…+k+(k+1) | (k+1)(k+2)/2 1+2+3+…+n = n(n+1)/2 Note that part of the left hand side in step #3 is the same as the left hand side in step #2. Previous Slide Slide 11 Next Slide**Step #1: Prove for n=1**1 | 1(1+1)/2 | 2/2 = 1 Step #2: Assume True for n=k 1+2+3+…+k = k(k+1)/2 Step #3: Using step #2, prove for n=k+1 1+2+3+…+(n-1)+n | n(n+1)/2 1+2+3+…+(k+1-1)+(k+1) | (k+1)(k+1+1)/2 1+2+3+…+k+(k+1) | (k+1)(k+2)/2 k(k+1)/2 +(k+1) | 1+2+3+…+n = n(n+1)/2 Now based on the equation in step #2 make a substitution. Previous Slide Slide 12 Next Slide**Step #1: Prove for n=1**1 | 1(1+1)/2 | 2/2 = 1 Step #2: Assume True for n=k 1+2+3+…+k = k(k+1)/2 Step #3: Using step #2, prove for n=k+1 1+2+3+…+(n-1)+n | n(n+1)/2 1+2+3+…+(k+1-1)+(k+1) | (k+1)(k+1+1)/2 1+2+3+…+k+(k+1) | (k+1)(k+2)/2 k(k+1)/2 +(k+1) | k(k+1)/2 +2(k+1)/2 | 1+2+3+…+n = n(n+1)/2 We now have to show that both sides are the same. This can be done in many ways. I’ll start by putting the second term on the left hand side over the LCD. Previous Slide Slide 13 Next Slide**Step #1: Prove for n=1**1 | 1(1+1)/2 | 2/2 = 1 Step #2: Assume True for n=k 1+2+3+…+k = k(k+1)/2 Step #3: Using step #2, prove for n=k+1 1+2+3+…+(n-1)+n | n(n+1)/2 1+2+3+…+(k+1-1)+(k+1) | (k+1)(k+1+1)/2 1+2+3+…+k+(k+1) | (k+1)(k+2)/2 k(k+1)/2 +(k+1) | k(k+1)/2 +2(k+1)/2 | [(k+1)/2] [ k +2 ] | 1+2+3+…+n = n(n+1)/2 Now I’ll factor out the common factor of (k+1)/2. Previous Slide Slide 14 Next Slide**Step #1: Prove for n=1**1 | 1(1+1)/2 | 2/2 = 1 Step #2: Assume True for n=k 1+2+3+…+k = k(k+1)/2 Step #3: Using step #2, prove for n=k+1 1+2+3+…+(n-1)+n | n(n+1)/2 1+2+3+…+(k+1-1)+(k+1) | (k+1)(k+1+1)/2 1+2+3+…+k+(k+1) | (k+1)(k+2)/2 k(k+1)/2 +(k+1) | k(k+1)/2 +2(k+1)/2 | [(k+1)/2] [ k +2 ] | (k+1)(k+2)/2 = 1+2+3+…+n = n(n+1)/2 Finally I’ll rewrite the left hand side to look like the right hand side. Previous Slide Slide 15 Next Slide**Step #1: Prove for n=1**1 | 1(1+1)/2 | 2/2 = 1 Step #2: Assume True for n=k 1+2+3+…+k = k(k+1)/2 Step #3: Using step #2, prove for n=k+1 1+2+3+…+(n-1)+n | n(n+1)/2 1+2+3+…+(k+1-1)+(k+1) | (k+1)(k+1+1)/2 1+2+3+…+k+(k+1) | (k+1)(k+2)/2 k(k+1)/2 +(k+1) | k(k+1)/2 +2(k+1)/2 | [(k+1)/2] [ k +2 ] | (k+1)(k+2)/2 = 1+2+3+…+n = n(n+1)/2 We now know that if the statement is true for any natural number, k, it also has to be true for the next natural number, k+1. Thus, since we know it’s true for 1, it also has to be true for 2. Since it’s true for 2, it has to be true for 3. And so on, and so on. Therefore, we have proven the statement is true for all natural numbers, 1,2,3,… . Previous Slide Slide 16 END