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Find the Pattern in 3+6+12+20+30 +n

  1. Nov 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Write a loop in C for to sum the following sequence 3+6+12+20+30+...nth.

    2. Relevant equations
    n/a

    3. The attempt at a solution
    I have tried to factor but there are no common factors between 3 and 20 other than 1.

    2*5=10------->10+20=30
    2*4=8--------->8+12=20
    2*3=6--------->6+6=12
    2*2=4--------->4+3=7:(

    I'm stuck. Any ideas?
     
  2. jcsd
  3. Nov 13, 2014 #2
    Recheck the first term. Hopefully it's 2.

    Otherwise OEIS is kind-of your friend - you'll maybe get an answer, but it probably won't be an answer you'll like. https://oeis.org/A066140 is not a sequence you can easily generate.

    The other (programming) option is to bounce the problem of the nth term into a function - which you leave unspecified.
     
  4. Nov 13, 2014 #3
    Thanks for the link. I have no idea what the reply was telling me.

    I'm pretty sure the 3 should be a 2 also.
     
  5. Nov 14, 2014 #4

    BvU

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    What if someone utters "hey this looks like ##\displaystyle 1+\sum_1^N n(n+1)\ ## "?
     
  6. Nov 14, 2014 #5
    If someone were to utter that, I would enthusiastically test values.

    N=4

    1+1(1+1) = 3
    1+2(2+1) = 7 o_O
    1+3(3+1) = 13 :(
    1+4(4+1) = 21 :L

    Then I would be disappointed and confused as to if I was doing the problem right.
     
  7. Nov 14, 2014 #6

    BvU

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    Don't understand. I see 3+6+12+20+30+...nth.
    I notice 3 = 1+ 1*2, 6 = 2*3, 12 = 3*4, 20 = 4*5, 30 = 5*6
    So methinks $$ 3+6+12+20+30 =
    \displaystyle 1+\sum_1^N n(n+1)\ $$

    (not ##\sum_1^N 1 + n(n+1)\ ## as in post #5)
     
  8. Nov 14, 2014 #7
    This is exactly the same as saying the first term of the series should be 2. - except it isn't, so lets just add a constant to that one term. You are also trying to jump to the sum instead of trying to identify the sequence.

    I thought of a kludge to get us around the "faulty" start to the sequence:

    ##s_n = \max(n+(n+1), n*(n+1))##
     
    Last edited: Nov 14, 2014
  9. Nov 14, 2014 #8

    BvU

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    Well, sorry for the confusion. I see your problem. My pattern recognition capabilities aren't up to having 3 as a starting term, so I made it (too?) easy for mself.
     
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