Mathematical Induction
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1) The document uses mathematical induction to prove several formulas. 2) It demonstrates proofs for formulas like 1 + 3 + 5 + ... + (2n-1) = n^2 and 2 + 4 + ... + 2n = n(n+1). 3) The proofs follow the standard structure of mathematical induction, showing the base case is true and using the induction hypothesis to show if the statement is true for n it is also true for n+1.
![Sk+1 = 1+3+5+7. . .+(2k–1)+[2(k+1)–1]
= [1+3+5+7+. . .+(2k–1)]+(2k+2–1)
= Sk+(2k+1)
= k2+2k+1
= (k+1)2](https://image.slidesharecdn.com/mathematicalinductionppt-130922102002-phpapp02/85/Mathematical-Induction-9-320.jpg)
Mathematical Induction
- 2. 1 + 3 + 5 + ... + (2n-1) = n2 1. Show it is true for n=1 1 = 12 is True 2. Assume it is true for n=k 1 + 3 + 5 + ... + (2k-1) = k2 is True Now, prove it is true for "k+1“ 1+3+5+...+(2k-1)+ 2(k+1)-1)=(k+1)2 ... ? We know that 1 + 3 + 5 + ... + (2k-1) = k2 (the assumption above), so we can do a replacement for all but the last term: k2 + (2(k+1)-1) = (k+1)2
- 3. Now expand all terms: k2 + 2k + 2 - 1 = k2 + 2k+1 And simplify: k2 + 2k + 1 = k2 + 2k + 1 They are the same! So it is true. So: 1 + 3 + 5 + ... + (2(k+1)- 1) = (k+1)2 is True
- 4. 1 + 2 + 3 + ... + n = n(n + 1) / 2 let S(n) be the statement 1+2+3+... n=n(n+1)/2 S(1) is the statement 1=1(1+1/2. Thus S(1) is true. We suppose that S(k) is true and prove that S(k + 1) is true. Thus, we assume that 1 + 2 + 3 + ... + k = k(k + 1) / 2 and prove that 1+2+3+...+k+k+1=(k+1)(k+1+1) / 2
- 5. If we add k + 1 to both sides of the equality in S(k), then on the left side of the sum, we obtain the left side of equality in S(k + 1). Our hope is that the right of the sum equals the right side of S(k + 1). Let us check: Adding K + 1 to both sides of S(k) we get: 1 + 2 + 3 + ... + k + k + 1 = k(k + 1) / 2 + (k + 1) = (k + 1)(k / 2 + 1) = (k + 1)(k / 2 + 2 / 2) = (k + 1)(k + 2) / 2 = (k + 1)(k + 1 + 1) / 2 Hence, if S(k) is true, then S(k + 1) is true. Therefore, 1 + 2 + 3 + ... + n = n(n + 1) / 2 for each positive integer n.
- 6. 1 + 3 + 5 + . . . + (2n − 1) = n2 for any integer n ≥ 1. Proof: STEP 1: For n=1 (1.2) is true, since 1 = 12 STEP 2: Suppose (1.2) is true for some n = k ≥ 1, that is 1 + 3 + 5 + . . . + (2k − 1) = k 2
- 7. STEP 3: Prove that (1.2) is true for n=k+1, that is 1+3+5+...+(2k−1)+(2k+1)?=(k+) 2 We have: 1+3+5+...+(2k−1)+(2k+1)=k 2+(2k+1)= (k + 1) 2
- 8. Sn = 1 + 3 + 5 + 7 + . . . + (2n-1) = n2 First, we must show that the formula works for n = 1. 1. For n = 1 S1 = 1 = 12 The second part of mathematical induction has two steps. The first step is to assume that the formula is valid for some integer k. The second step is to use this assumption to prove that the formula is valid for the next integer, k + 1. 2. Assume Sk = 1 + 3 + 5 + 7 + . . . + (2k-1) = k2 is true, show that Sk+1 = (k + 1)2 is true.
- 9. Sk+1 = 1+3+5+7. . .+(2k–1)+[2(k+1)–1] = [1+3+5+7+. . .+(2k–1)]+(2k+2–1) = Sk+(2k+1) = k2+2k+1 = (k+1)2
- 10. 12 + 22 + ... + n2 = n( n + 1 )( 2n + 1 )/6 If n = 0, then LHS = 02 = 0, and RHS = 0 * (0 + 1)(2*0 + 1)/6 = 0 Hence LHS = RHS. 12 + 22 + ... + n2 = n( n + 1 )( 2n + 1 )/6 --- Induction Hypothesis To prove this for n+1, first try to express LHS for n+1 in terms of LHS for n, and use the induction hypothesis. Here let us try LHS for n + 1 = 12 + 22 + ... + n2 + (n + 1)2 = ( 12 + 22 + ... + n2 ) + (n + 1)2
- 11. Using the induction hypothesis, the last expression can be rewritten as n( n + 1 )( 2n + 1 )/6 + (n + 1)2 Factoring (n + 1)/6 out, we get ( n + 1 )( n( 2n + 1 ) + 6 ( n + 1 ) )/6 = ( n + 1 )( 2n2 + 7n + 6 )/6 = ( n + 1 )( n + 2 )( 2n + 3 )/6 , which is equal to the RHS for n+1. Thus LHS = RHS for n+1.
- 12. n , n3 + 2n is divisible by 3. If n = 0, then n3 + 2n = 03 + 2*0 = 0. So it is divisible by 3. Induction: Assume that for an arbitrary natural number n, n3 + 2n is divisible by 3. --- Induction Hypothesis To prove this for n+1, first try to express (n+1)3 + 2( n + 1 ) in terms of n3 + 2n and use the induction hypothesis.
- 13. (n+1)3+2(n+1)=(n3+3n2+3n+1)+( 2n + 2 ) = ( n3 + 2n ) + ( 3n2 + 3n + 3 ) = ( n3 + 2n ) + 3( n2 + n + 1 ) which is divisible by 3, because ( n3 + 2n ) is divisible by 3 by the induction hypothesis.
- 14. 2 + 4 + ... + 2n = n( n + 1 ) If n=0, then LHS=0, and RHS=0*(0+1)=0 Hence LHS = RHS Assume that for an arbitrary natural number n, 0+2+...+2n=n(n+1) --- Induction Hypothesis To prove this for n+1, first try to express LHS for n+1 in terms of LHS for n, and somehow use the induction hypothesis.
- 15. Here let us try LHS for n+1=0+2+...+2n+2(n+1)=(0+2+ ...+2n)+2(n+1) Using the induction hypothesis, the last expression can be rewritten as n( n + 1 ) + 2(n + 1) Factoring (n + 1) out, we get (n + 1)(n + 2) , which is equal to the RHS for n+1. Thus LHS = RHS for n+1.