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Homework Help: Help with math Progressions

  1. May 8, 2005 #1
    (1) Lets say you have the progression [itex] 5 + 12 + 21 + ... + 1048675 [/itex] and you want to find the sum of 20 terms. I know that the kth term is given by [itex] 2^{k} +3 + 5(k-1) [/itex]. So would I treat the [itex] 2^{k} [/itex] terms separately from the [itex] 3 + 5(k-1) [/itex] terms? Would it be [itex] 2 + 4 + 8 + 16 +... + 2^{n} [/itex] and [itex] 3 + 8 + 13 + ... + (3+5(k-1)) [/itex]. Would the total sum be [itex] \frac{2 - 2(2)^{n}}{-1} + \frac{n}{2}(3+ (3+5(k-1)) [/itex]?

    [tex] 2097248 + 1010 [/tex]

    (2) [tex] 3 + 10 + 25 + ... + 39394 [/tex] and you want to find sum of first 10 terms. I know that the kth term is [itex] 2 \times 3^{k-1} + 1 + 3(k-1) [/itex] Would i do the same thing and treat the [itex] 3^{k-1} [/itex] and [itex] 1 + 3(k-1) [/itex] separately?

    Thanks
     
    Last edited: May 8, 2005
  2. jcsd
  3. May 8, 2005 #2
    I have only looked at (1) but here is what I have done:

    a1 = 5
    a2 = 12
    a3 = 21

    You have said that: ak = 2k + 5k - 2

    So: ak + 1 = 2(k+1) + 5(k+1) - 2

    This means that: ak+1 + ak = 2(k+1) + 5(k+1) - 2 + 2k + 5k - 2 = 2k + 1 + 2k + 10k + 1

    and: This means that: ak+2 + ak+1 + ak = 2(k+2) + 5(k+2) - 2 + 2(k+1) + 5(k+1) - 2 + 2k + 5k - 2
    = 2(k+2) + 2(k+1) + 2k + 15k + 9

    I have found that the sum of the terms 2(k+n) for n = 0, 1, 2 etc. is 2(k+1) - 2.

    The term for the sum of the 5k, 10k, 15k terms is equal to 5k2

    However I was unable to work out the last term. All you need to do, from what I have done, is work out the way to express the difference between -2, 1 and 9 in terms of k when k is 1, 2 and 3 respectively. Then you can put all three terms together and you can find the sum up to any kth term.

    The Bob (2004 ©)
     
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