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Find the nth term

  1. Apr 17, 2005 #1
    I've got a very tricky question on my hands.

    A set of integers are grouped as follows

    [tex] (1), (2,3,4), (5,6,7,8,9),..., [/tex] until the nth bracket.

    I have found the total integers in the first (n-1) brackets and it is [tex] (n-1)^2 [/tex]. The next part of the question is to show that the first number in the first term in the nth bracket is [tex] n^2-2n+2 [/tex]. What i did was to first write out the sequence representing the first term in each bracket,

    [tex] 1,2,5,10,17,...[/tex]

    but i can't seem to find any pattern with this sequence but have only seen that their difference is an arithmetic progression. How do I go about this question?
     
  2. jcsd
  3. Apr 17, 2005 #2

    quasar987

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    You hardly have to show this. There are (n-1)² terms contained in the first (n-1) brackets. So the LAST term of the (n-1)th bracket is the number (n-1)². This makes the first term of the nth bracket (n-1)² + 1 = n² - 2n +2
     
  4. Apr 17, 2005 #3

    dextercioby

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    And that sequence of #-s [itex] 1,2,5,10,17,26,... [/itex] can be described by

    [tex] a_{n+1}=a_{n}+2n-1 [/tex]

    Daniel.
     
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