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mwmnj

  • 3 years ago

Proof help? http://imgur.com/bqFCa

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  1. mwmnj
    • 3 years ago
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    So first step of mathematical induction is show the first case is true, so i plug in 1 for i on the left side and n on the right side:

  2. neverforgetvivistee
    • 3 years ago
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    lol that's from reddit

  3. mwmnj
    • 3 years ago
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    \[\sum_{1}^{i=1}(1)2^{(1)}=\]

  4. mwmnj
    • 3 years ago
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    Whats from reddit?

  5. mwmnj
    • 3 years ago
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    So above, on the left i have 2

  6. mwmnj
    • 3 years ago
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    on the right....

  7. strobe
    • 3 years ago
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    It's the sum up to n + 1, not n

  8. mwmnj
    • 3 years ago
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    \[(1)2^{((1)+2)}+2 = 2^{3}+2 = 10\]

  9. mwmnj
    • 3 years ago
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    @ strobe, so i am supposed to plug in a 2 instead of a 1?

  10. strobe
    • 3 years ago
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    It's the sum of the LHS subbed with 1 and 2

  11. mwmnj
    • 3 years ago
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    LHS?

  12. mwmnj
    • 3 years ago
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    left hand side

  13. strobe
    • 3 years ago
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    Yah

  14. mwmnj
    • 3 years ago
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    ok so the RHS is right

  15. mwmnj
    • 3 years ago
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    \[\sum_{i = 2}^{2} = 2(2)^{2} = 8\]

  16. mwmnj
    • 3 years ago
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    so add that to the first summation and i get 10 on the LHS

  17. mwmnj
    • 3 years ago
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    so 10 = 10 is true

  18. mwmnj
    • 3 years ago
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    Is that correct?

  19. strobe
    • 3 years ago
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    Yes

  20. mwmnj
    • 3 years ago
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    Ok, thanks, ill see if I can finish the rest from there :)

  21. strobe
    • 3 years ago
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    Good luck!

  22. mwmnj
    • 3 years ago
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    after step one (show p(1) is true: step 2 is assume p(k) is true and then show p(k) implies p(k+1)

  23. mwmnj
    • 3 years ago
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    Cant quite see how i can make that happen at the moment, heres what I have:

  24. mwmnj
    • 3 years ago
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    \[p(k) : \sum_{i=1}^{k+1}i2^{i} = k2^{k+2}+2\]

  25. mwmnj
    • 3 years ago
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    \[p(k+1): \sum_{i=1}^{k+2}i2^{i}=(k+1)2^{k+3}+2\]

  26. mwmnj
    • 3 years ago
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    RHS simplifies to \[k2^{k+3}+2^{k+3}+2\]

  27. mwmnj
    • 3 years ago
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    So now I need to add something to p(k) to make it match p(k+1)

  28. mwmnj
    • 3 years ago
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    and if after adding that to p(k) the RHS of p(k) = the RHS of p(k+1), that is sufficient proof

  29. strobe
    • 3 years ago
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    I wouldn't bother expanding the RHS. I would expand the LHS, then put it in terms of p(k) + something, replace p(k) with the LHS, then rearrange that result so it looks like the LHS of p(k+1)

  30. mwmnj
    • 3 years ago
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    \[p(k)+(k+1)2^{k+1}: \sum_{i=1}^{k+1}i2^{i}+(k+1)2^{k+1}=k2^{k+2}+2+(k+1)2^{k+1}\]

  31. mwmnj
    • 3 years ago
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    Hmm, i was always taught to modify add to p(k) then modify p(k)+ whateverRHS to make it the same as RHS of p(k+1)

  32. mwmnj
    • 3 years ago
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    This is so freakin complicated

  33. mwmnj
    • 3 years ago
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    "I would calculate the RHS for k + 1 and then leave it alone. Then try and change the LHS to the same form as the RHS but substituting in the the RHS of p(k) into the LHS."

  34. mwmnj
    • 3 years ago
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    So first I should calculate the right hand side of p(k+1)?

  35. mwmnj
    • 3 years ago
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    or of p(k)

  36. strobe
    • 3 years ago
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    p(k+1), but dont expand or simplify it, leave it as it is. This is what you're aiming for with the LHS.

  37. mwmnj
    • 3 years ago
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    the LHS of p(k+1)

  38. mwmnj
    • 3 years ago
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    ok so,

  39. mwmnj
    • 3 years ago
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    RHS of p(k+1): \[(k+1)2^{(k+1)+2}+2\]

  40. mwmnj
    • 3 years ago
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    isnt much calculation to do really, just plug in

  41. mwmnj
    • 3 years ago
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    Then I want to make LHS of P(k+1) the same as that:

  42. mwmnj
    • 3 years ago
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    RHS of p(k+1): \[\sum_{i=1}^{(k+1)+1}i2^{i}\]

  43. mwmnj
    • 3 years ago
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    *LHS rather

  44. strobe
    • 3 years ago
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    Yup, so that equals the LHS of p(k) + something, find that something

  45. mwmnj
    • 3 years ago
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    So I should expand the LHS of p(k+1) first?

  46. strobe
    • 3 years ago
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    Expand so it looks like LHS of p(k) + something

  47. mwmnj
    • 3 years ago
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    I would need to add (k+1)2^(k+1) to LHS of p(k) for them to be equal

  48. mwmnj
    • 3 years ago
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    I just have to show that...

  49. strobe
    • 3 years ago
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    Okay, so LHS p(k+1) = LHS p(k) + (k+2)2^(k+2) (you almost got it) but we know LHS = RHS of p(k), so substitute the LHS with the RHS

  50. mwmnj
    • 3 years ago
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    \[(p(k)LHS)+(k+1)2 ^{k+1}-> (k2^{k+2}+2)+(k+1)2 ^{k+1}\]

  51. mwmnj
    • 3 years ago
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    The substitution into (p(k)LHS) is p(k)RHS

  52. mwmnj
    • 3 years ago
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    Now I need to make that = the RHS or p(k+1)?

  53. mwmnj
    • 3 years ago
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    If so, that's the method I am familiar with

  54. strobe
    • 3 years ago
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    Yup, you'll find it all factorises nicely

  55. mwmnj
    • 3 years ago
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    ok, I must have bliped up somewhere before then, cause the Q i asked making the sides equal was my next step from here attempted

  56. mwmnj
    • 3 years ago
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    I'll try it here

  57. mwmnj
    • 3 years ago
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    so im trying to make \[(k2^{k+2}+2)+(k+1)2^{k+1} = (k+1)2^{(k+1)+2}+2 \]

  58. mwmnj
    • 3 years ago
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    expanding everything out isnt the way to go?

  59. neverforgetvivistee
    • 3 years ago
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    yes

  60. mwmnj
    • 3 years ago
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    it is the way to go?

  61. mwmnj
    • 3 years ago
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    hmm

  62. mwmnj
    • 3 years ago
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    p(k): \[k2^{k}4+2^{k}2+4\]

  63. mwmnj
    • 3 years ago
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    Im not seeing it :/

  64. mwmnj
    • 3 years ago
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    Clearly, im totally lost

  65. phi
    • 3 years ago
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    You have to be very careful with the indices with this problem. We can show the expression works for n=1. Now assume it is true for n=m. Can we show it is true for m+1?

  66. mwmnj
    • 3 years ago
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    Thats what i tried to show

  67. mwmnj
    • 3 years ago
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    Did i make an arithmetic error somewhere or is my methodology wrong or what

  68. phi
    • 3 years ago
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    \[\sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)w^{m+2}\]

  69. phi
    • 3 years ago
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    not sure where you went wrong. See how we start: We increased m by 1, and then break the sum into two parts.

  70. phi
    • 3 years ago
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    the first sum on the right hand side "works" because we assume it works for n=m. so we can replace it with \[ m 2^{m+2} +2 \]

  71. phi
    • 3 years ago
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    that "w" in the first formula is a 2 ! \[ \sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)2^{m+2}\] \[ = 2+ m 2^{m+2} +(m+2)2^{m+2} \]

  72. mwmnj
    • 3 years ago
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    ok

  73. mwmnj
    • 3 years ago
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    so now I need to set that equal to the right hand side of m+1

  74. mwmnj
    • 3 years ago
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    or make it equal that at least

  75. phi
    • 3 years ago
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    I'm still working on it. we want the right hand side equal to what the formula says if we use n= m+1

  76. mwmnj
    • 3 years ago
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    RHS of m+1:\[(m+1)2^{m+3}+2 \]

  77. phi
    • 3 years ago
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    that "w" in the first formula is a 2 ! \[ \sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)2^{m+2}\] \[ = 2+ m 2^{m+2} +(m+2)2^{m+2} \]

  78. mwmnj
    • 3 years ago
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    right we need to make that equal what i just typed up right?

  79. mwmnj
    • 3 years ago
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    If so, that is this: http://imgur.com/P8Tnf

  80. phi
    • 3 years ago
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    yes and I believe they are.

  81. mwmnj
    • 3 years ago
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    Just with m

  82. mwmnj
    • 3 years ago
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    isnt it or am i missing something?

  83. phi
    • 3 years ago
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    we have m+2 now, you have k+1

  84. phi
    • 3 years ago
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    very close, but not close enough!

  85. mwmnj
    • 3 years ago
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    ok i must have messed that up somewhere

  86. phi
    • 3 years ago
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    Do you see how to get the "new" version? (with m+2), it's the next term we are adding, with i= (m+2)

  87. phi
    • 3 years ago
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    It's confusing because we go to n+1 instead of n, so going up one term means n+2. Anyways... can you simplify what we have?

  88. mwmnj
    • 3 years ago
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    yea, so confusing!

  89. phi
    • 3 years ago
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    multiply out (m+2) 2^(m+2)

  90. mwmnj
    • 3 years ago
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    making it m made it less tho

  91. mwmnj
    • 3 years ago
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    lets see

  92. phi
    • 3 years ago
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    we get \[ 2+ m 2^{m+2}+m2^{m+2} + 2^{m+3} \]

  93. mwmnj
    • 3 years ago
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    on LHS right

  94. mwmnj
    • 3 years ago
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    thats not the same tho

  95. mwmnj
    • 3 years ago
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    thats \[2+2m ^{m+2}+2^{m+3}\]

  96. mwmnj
    • 3 years ago
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    on the LHS

  97. phi
    • 3 years ago
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    add the two m2^(m+2) to get m*2*2^(m+2) = m*2^(m+3)

  98. mwmnj
    • 3 years ago
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    RHS is \[2+m2^{m+3}+2^{m+3}+2\]

  99. mwmnj
    • 3 years ago
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    Ah yes

  100. mwmnj
    • 3 years ago
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    Now I see it :)

  101. phi
    • 3 years ago
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    you have an extra 2 in there. But factor out 2^(m+3) to get \[ 2+ (m+1) 2^{m+3} \]

  102. mwmnj
    • 3 years ago
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    Let me try and write this out and think it out

  103. mwmnj
    • 3 years ago
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    hopefully fully see where i bliped up

  104. phi
    • 3 years ago
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    ok have fun

  105. mwmnj
    • 3 years ago
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    Thanks so much

  106. mwmnj
    • 3 years ago
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    Will post here when done

  107. mwmnj
    • 3 years ago
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    http://imgur.com/BYB4E

  108. phi
    • 3 years ago
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    I would tweak it a bit. Your statement showing that the formula works for p(1) is not clear. It is better to state: by definition \[ p(1)=\sum_{1}^{2} i2^i= 1\cdot2^1+2\cdot 2^2= 10\] From the formula, p(1) is \[ p(1)= 1 \cdot 2^{(1+2)}+2=10 \] therefore the formula holds for n=1 Also add a short explanation of what you are doing in part 2b. For example, The formula gives p(k+1) as .... and for the second part: By definition, p(k+1)= .... Finally, point out where you are making use of the assumption that the formula is assumed to be true for p(k). This is where you replace the summation with the results of the formula.

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