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mwmnj Group Title

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

  • 2 years ago
  • 2 years ago

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  1. mwmnj Group Title
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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 years ago
  2. neverforgetvivistee Group Title
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    lol that's from reddit

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

    • 2 years ago
  4. mwmnj Group Title
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    Whats from reddit?

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

    • 2 years ago
  6. mwmnj Group Title
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    on the right....

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

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

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

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

    • 2 years ago
  11. mwmnj Group Title
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    LHS?

    • 2 years ago
  12. mwmnj Group Title
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    left hand side

    • 2 years ago
  13. strobe Group Title
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    Yah

    • 2 years ago
  14. mwmnj Group Title
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    ok so the RHS is right

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

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

    • 2 years ago
  17. mwmnj Group Title
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    so 10 = 10 is true

    • 2 years ago
  18. mwmnj Group Title
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    Is that correct?

    • 2 years ago
  19. strobe Group Title
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    Yes

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

    • 2 years ago
  21. strobe Group Title
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    Good luck!

    • 2 years ago
  22. mwmnj Group Title
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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)

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

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

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

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

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

    • 2 years ago
  28. mwmnj Group Title
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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

    • 2 years ago
  29. strobe Group Title
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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)

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

    • 2 years ago
  31. mwmnj Group Title
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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)

    • 2 years ago
  32. mwmnj Group Title
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    This is so freakin complicated

    • 2 years ago
  33. mwmnj Group Title
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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."

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

    • 2 years ago
  35. mwmnj Group Title
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    or of p(k)

    • 2 years ago
  36. strobe Group Title
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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.

    • 2 years ago
  37. mwmnj Group Title
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    the LHS of p(k+1)

    • 2 years ago
  38. mwmnj Group Title
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    ok so,

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

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

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

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

    • 2 years ago
  43. mwmnj Group Title
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    *LHS rather

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

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

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

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

    • 2 years ago
  48. mwmnj Group Title
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    I just have to show that...

    • 2 years ago
  49. strobe Group Title
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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

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

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

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

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

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

    • 2 years ago
  55. mwmnj Group Title
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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

    • 2 years ago
  56. mwmnj Group Title
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    I'll try it here

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

    • 2 years ago
  58. mwmnj Group Title
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    expanding everything out isnt the way to go?

    • 2 years ago
  59. neverforgetvivistee Group Title
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    yes

    • 2 years ago
  60. mwmnj Group Title
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    it is the way to go?

    • 2 years ago
  61. mwmnj Group Title
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    hmm

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

    • 2 years ago
  63. mwmnj Group Title
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    Im not seeing it :/

    • 2 years ago
  64. mwmnj Group Title
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    Clearly, im totally lost

    • 2 years ago
  65. phi Group Title
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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?

    • 2 years ago
  66. mwmnj Group Title
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    Thats what i tried to show

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

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

    • 2 years ago
  69. phi Group Title
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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.

    • 2 years ago
  70. phi Group Title
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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 \[ m 2^{m+2} +2 \]

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

    • 2 years ago
  72. mwmnj Group Title
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    ok

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

    • 2 years ago
  74. mwmnj Group Title
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    or make it equal that at least

    • 2 years ago
  75. phi Group Title
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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

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

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

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

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

    • 2 years ago
  80. phi Group Title
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    yes and I believe they are.

    • 2 years ago
  81. mwmnj Group Title
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    Just with m

    • 2 years ago
  82. mwmnj Group Title
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    isnt it or am i missing something?

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

    • 2 years ago
  84. phi Group Title
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    very close, but not close enough!

    • 2 years ago
  85. mwmnj Group Title
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    ok i must have messed that up somewhere

    • 2 years ago
  86. phi Group Title
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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)

    • 2 years ago
  87. phi Group Title
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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?

    • 2 years ago
  88. mwmnj Group Title
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    yea, so confusing!

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

    • 2 years ago
  90. mwmnj Group Title
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    making it m made it less tho

    • 2 years ago
  91. mwmnj Group Title
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    lets see

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

    • 2 years ago
  93. mwmnj Group Title
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    on LHS right

    • 2 years ago
  94. mwmnj Group Title
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    thats not the same tho

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

    • 2 years ago
  96. mwmnj Group Title
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    on the LHS

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

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

    • 2 years ago
  99. mwmnj Group Title
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    Ah yes

    • 2 years ago
  100. mwmnj Group Title
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    Now I see it :)

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

    • 2 years ago
  102. mwmnj Group Title
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    Let me try and write this out and think it out

    • 2 years ago
  103. mwmnj Group Title
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    hopefully fully see where i bliped up

    • 2 years ago
  104. phi Group Title
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    ok have fun

    • 2 years ago
  105. mwmnj Group Title
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    Thanks so much

    • 2 years ago
  106. mwmnj Group Title
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    Will post here when done

    • 2 years ago
  107. mwmnj Group Title
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    http://imgur.com/BYB4E

    • 2 years ago
  108. phi Group Title
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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.

    • 2 years ago
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