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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.

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  1. aussy123
    • 2 years ago
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  2. perl
    • 2 years ago
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    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
    • 2 years ago
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    Im following

  4. perl
    • 2 years ago
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    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
    • 2 years ago
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    okay

  6. perl
    • 2 years ago
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    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
    • 2 years ago
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    this is the principle of mathematical induction. I didnt actually do your example yet, this just preliminary

  8. aussy123
    • 2 years ago
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    I know, thank you, good notes

  9. perl
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    so we can rewrite the statement P(n)

  13. aussy123
    • 2 years ago
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    ok im starting to understand

  14. perl
    • 2 years ago
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    |dw:1366641535256:dw|

  15. perl
    • 2 years ago
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    so we have here a left side and a right side of the statement we are trying to prove

  16. perl
    • 2 years ago
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    P(1) : |dw:1366641602052:dw|

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

  18. aussy123
    • 2 years ago
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    I thought p(1) was 1(1+2)

  19. perl
    • 2 years ago
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    we want to prove [2(1) +1] + [2(2) + 1] + [2(3)+1] + ... [2(n)+1] = n(n+1)

  20. aussy123
    • 2 years ago
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    oh

  21. perl
    • 2 years ago
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    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
    • 2 years ago
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    yes

  23. perl
    • 2 years ago
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    because 3 = 2(1) + 1 5 = 2 (2) + 1 7 = 2(3) + 1

  24. perl
    • 2 years ago
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    |dw:1366641946782:dw|

  25. perl
    • 2 years ago
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    let's look at the case n=2 |dw:1366642018761:dw|

  26. perl
    • 2 years ago
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    |dw:1366642104889:dw|

  27. perl
    • 2 years ago
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    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
    • 2 years ago
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    yes, but which number is k because in my problem k isnt named

  29. perl
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    oh dear i didnt copy correctly

  32. perl
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    3=3

  35. perl
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    with me so far?

  39. perl
    • 2 years ago
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    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
    • 2 years ago
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    yeah

  41. perl
    • 2 years ago
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    using only algebra, basically

  42. perl
    • 2 years ago
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    so how about we look at the conclusion for a minute

  43. perl
    • 2 years ago
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    ok, let me do it another way, we can go back to this way later, it might be a little advanced

  44. perl
    • 2 years ago
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    ok still with me?

  45. aussy123
    • 2 years ago
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    yea Im following

  46. perl
    • 2 years ago
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    we assume that [2(1) + 1 ] + [2(2)+1 ] + . . . + [ 2(k)+ 1 ] = k ( k + 2) is true

  47. perl
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    |dw:1366643507200:dw|

  50. perl
    • 2 years ago
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    so we are just using algebra here

  51. aussy123
    • 2 years ago
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    ok

  52. perl
    • 2 years ago
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    you can add the same number to both sides of an equation, thats a legal algebraic move dfsd

  53. perl
    • 2 years ago
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    ok im back , :)

  54. aussy123
    • 2 years ago
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    ok

  55. perl
    • 2 years ago
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    so we added [2(k+1) + 1 ] to both sides of the hypothesis , which was given

  56. aussy123
    • 2 years ago
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    hello?

  57. perl
    • 2 years ago
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    yes did you add it?

  58. perl
    • 2 years ago
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    |dw:1366644222445:dw|

  59. perl
    • 2 years ago
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    now the left side is what we want in our conclusion, and we can simplify the right side

  60. aussy123
    • 2 years ago
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    okay

  61. aussy123
    • 2 years ago
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    Im here but for some reason I keep losing connection

  62. perl
    • 2 years ago
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    me too

  63. perl
    • 2 years ago
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    i was lagged

  64. aussy123
    • 2 years ago
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    me too

  65. perl
    • 2 years ago
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    ok did you add both sides

  66. aussy123
    • 2 years ago
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    I tried but my teacher said we already solved for what was needed.

  67. perl
    • 2 years ago
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    [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

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