aussy123 2 years ago PLEASE! IM BEGGING!! HELP ME!! PLEASE!! *TEARS* Can someone please help me with this? I post it earlier but the answer wasnt what my teacher wanted. Can someone help me I will post in attachment.

1. aussy123

2. perl

we want to prove that the formula 3 + 5 + 7 + ... + 2n+1 = n(n+1) is true for positive integers n=1,2,3,... the problem is, there are an infinite number of statements to check, and we will never get done, one for each number n. so we use mathematical induction

3. aussy123

Im following

4. perl

I am going to rewrite this as a proposition P(n) Let P(n) be the statement 3 + 5 + 7 + ... + 2n+1 = n(n+1)

5. aussy123

okay

6. perl

mathematical induction says : If you have a statement that is true for some number, for example n=1 (this is called the basis case) and furthermore that it can be proven if P(k) is true, then P(k+1) is true. then it must be the case that P(n) is true for all n greater than your basis

7. perl

this is the principle of mathematical induction. I didnt actually do your example yet, this just preliminary

8. aussy123

I know, thank you, good notes

9. perl

so for instance if your basis case P(1), and assuming you showed P(k)->P(k+1) then you have P(2) because P(1) is true and P(1) ->P(2) is true then it must be true P(2) is true , etc

10. perl

ok thats just the logic behind it. so in our example P(n) : 3 + 5 + 7 + ... + 2n+1 = n(n+1) is P(1) true?

11. perl

notice on the left side, you are summing a sequence of odd numbers 3,5,7,...The formula for the sequence of odd numbers is 2n+1 ,

12. perl

so we can rewrite the statement P(n)

13. aussy123

ok im starting to understand

14. perl

|dw:1366641535256:dw|

15. perl

so we have here a left side and a right side of the statement we are trying to prove

16. perl

P(1) : |dw:1366641602052:dw|

17. perl

now the left side is just another way of saying, plug in 1 into the formula 2k+1 , no big deal.

18. aussy123

I thought p(1) was 1(1+2)

19. perl

we want to prove [2(1) +1] + [2(2) + 1] + [2(3)+1] + ... [2(n)+1] = n(n+1)

20. aussy123

oh

21. perl

ok let me start over, we want to prove this statement is true for all n. [2(1) +1] + [2(2) + 1] + [2(3)+1] + ... [2(n)+1] = n(n+1) do you agree?

22. aussy123

yes

23. perl

because 3 = 2(1) + 1 5 = 2 (2) + 1 7 = 2(3) + 1

24. perl

|dw:1366641946782:dw|

25. perl

let's look at the case n=2 |dw:1366642018761:dw|

26. perl

|dw:1366642104889:dw|

27. perl

case n=1 2(1) + 1 = 1 (1+1) case n=2 [ 2(1) + 1 ] + [2 (2) + 1 ] = 2 (2+1) case n=3 [ 2(1) + 1 ] + [2 (2) + 1 ] + [ 2(3) + 1] = 3 (3+1) do you start to see a pattern?

28. aussy123

yes, but which number is k because in my problem k isnt named

29. perl

we want to prove all these statements are true without actually have to check each one by hand or calculator [2(1) + 1] = 1 (1+1) [ 2(1) + 1 ] + [2 (2) + 1 ] = 2 (2+1) [ 2(1) + 1 ] + [2 (2) + 1 ] + [ 2(3) + 1] = 3 (3+1) [ 2(1) + 1 ] + [2 (2) + 1 ] + [ 2(3) + 1] +[2(4) + 1] = 4 (4+1) . . . [2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(n)+ 1 ] = n ( n + 1)

30. perl

the way to prove all of the statements, first show it is true for the first one [2(1) + 1] = 1 (1+1) , does the left side equal right side (yes it does). thats the basis case

31. perl

oh dear i didnt copy correctly

32. perl

Correction we want to prove all these statements are true for any number n without actually have to check each one by hand or calculator [2(1) + 1] = 1 (1+2) [ 2(1) + 1 ] + [2 (2) + 1 ] = 2 (2+2) [ 2(1) + 1 ] + [2 (2) + 1 ] + [ 2(3) + 1] = 3 (3+2) [ 2(1) + 1 ] + [2 (2) + 1 ] + [ 2(3) + 1] +[2(4) + 1] = 4 (4+2) . . . [2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(n)+ 1 ] = n ( n + 2)

33. perl

Proof: first show it is true for the first case [2(1) + 1] = 1 (1+2) ( Check to see if the left side equals right side (yes it does). This is called the basis case.

34. aussy123

3=3

35. perl

next step, (called the inductive step) show that if the statement is true for n=k, then it will be true for n=k+1

36. perl

so we prove that if this is true : [2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k)+ 1 ] = k ( k + 2) Then this must be true [2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k+1)+ 1 ] = (k+1) [ (k+1) + 2]

37. perl

i didnt do the proof yet , just showing you what we want to prove. we are proving what is called a 'conditional' statement. an 'if-then' statement

38. perl

with me so far?

39. perl

so we are given the 'if' part of the condition, also called the inductive hypothesis. so we assume that it is true that [2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k)+ 1 ] = k ( k + 2) now how can we go from this statement to this statement [2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k+1)+ 1 ] = (k+1) [ (k+1) + 2]

40. aussy123

yeah

41. perl

using only algebra, basically

42. perl

so how about we look at the conclusion for a minute

43. perl

ok, let me do it another way, we can go back to this way later, it might be a little advanced

44. perl

ok still with me?

45. aussy123

yea Im following

46. perl

we assume that [2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k)+ 1 ] = k ( k + 2) is true

47. perl

from this hypothesis we want to deduce the conclusion [2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k+1)+ 1 ] = (k+1) [ (k+1) + 2]

48. perl

proof: given [2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k)+ 1 ] = k ( k + 2) add [2(k+1) + 1 ] to both sides of equation

49. perl

|dw:1366643507200:dw|

50. perl

so we are just using algebra here

51. aussy123

ok

52. perl

you can add the same number to both sides of an equation, thats a legal algebraic move dfsd

53. perl

ok im back , :)

54. aussy123

ok

55. perl

so we added [2(k+1) + 1 ] to both sides of the hypothesis , which was given

56. aussy123

hello?

57. perl

58. perl

|dw:1366644222445:dw|

59. perl

now the left side is what we want in our conclusion, and we can simplify the right side

60. aussy123

okay

61. aussy123

Im here but for some reason I keep losing connection

62. perl

me too

63. perl

i was lagged

64. aussy123

me too

65. perl

ok did you add both sides

66. aussy123

I tried but my teacher said we already solved for what was needed.

67. perl

[2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k)+ 1 ] = k ( k + 2) +[2(k+1) +1 ] + [2(k+1)+1] --------------------------------------------------------- [2(1)+1] +[2(2)+1] +... +[2(k)+1] +[2(k+1)+1] = k(k+2) + [2(k+1)+1