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Sequence problem

  1. Jan 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider the sequence in the form of groups (1), (2,2), (3,3,3), (4,4,4,4), (5,5,5,5,5),...

    i. The 2000th term of the sequence is not divisible by
    (A) 3 (B) 9 (C) 7 (D) None of these

    ii. The sum of the first 2000 terms is
    (A) 84336 (B) 96324 (C) 78466 (D)None of these

    iii. The sum of the remaining terms in the group after 2000th term in which 2000th term lies is
    (A) 1088 (B) 1008 (C) 1040 (D) None of these

    2. Relevant equations


    3. The attempt at a solution
    I don't know where to start!!
    For ii, I need to find 1^2 + 2^2 + 3^2 + 4^2..... for which I have the formula n(n+1)(2n+1)/6, but I don't know what the preceding number of the group in which the 2000th term lies will be, so I can't even approximate an answer.
    Just please tell me how to begin!
     
  2. jcsd
  3. Jan 20, 2015 #2

    haruspex

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    I need a bit more background. In what sense is (5,5,5,5,5) a group, and in what sense might it be divisible by an integer?
     
  4. Jan 20, 2015 #3
    1-first term 2-second term 2-third term 3-fourth term 3-fifth term 3-sixth term 4- seventh term
    If the questions were-
    The sixth term of the sequence is not divisible by-
    (A)1 (B)2 (C)3
    Then the answer would be (B) as 3 is not divisible by 2.

    Sum of the first six terms-
    Ans: 14

    The sum of the remaining terms in the group after 6th term in which 6th term lies is
    Ans: 0
    OR
    The sum of the remaining terms in the group after 5th term in which 5th term lies is
    Ans: 3
     
  5. Jan 20, 2015 #4
    Sorry for the repeat post.

    Moderator note: It's now deleted.
     
    Last edited by a moderator: Jan 20, 2015
  6. Jan 20, 2015 #5

    haruspex

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    Ok, I see.
    Not sure why you think you want the formula for the sum of consecutive squares. You are not trying to sum the series.
    How many terms are there before the occurrence of the first n?
     
  7. Jan 20, 2015 #6
    Before the occurrence of the first n??
     
  8. Jan 20, 2015 #7

    haruspex

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    As in, zero before the first 1, one before the first 2, three before the first 3, six before the first 4,.... How many before the first n?
     
  9. Jan 20, 2015 #8
    n(n-1)/2 before the first n. What now?
     
  10. Jan 20, 2015 #9

    haruspex

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    So if the 2000th term is n, what equation can you write down?
     
  11. Jan 22, 2015 #10
    n(n-1)/2 < = 2000 ?
     
  12. Jan 22, 2015 #11
    Oh ok!!! n=63 gives 1953 numbers before 63. I got all the answers! Thank you so much :D
     
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