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Function is F(n) = n(n+1)

  1. Oct 15, 2008 #1
    F(n) = 2+4+6=...+2n
    I know the expression that represents the given function is F(n) = n(n+1),

    my question is F(n) = 3+5+7...2n the same F(n) = n(n+1)< if not can anyone expain?

    Thanks
     
  2. jcsd
  3. Oct 15, 2008 #2

    mathman

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    Re: Recursion

    Please correct! You have a sequence of odd numbers, but the end term is even!
     
  4. Oct 16, 2008 #3
    Re: Recursion

    This was gone into from General Math: Simple Sequences. The nth term is

    [tex] \sum_1^n (2j-1)=n^2. [/tex] Or: 1=1, 1+3=2^2, 1+3+5 = 3^2, etc...
     
    Last edited: Oct 16, 2008
  5. Oct 16, 2008 #4

    HallsofIvy

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    Re: Recursion

    Why should they have the same sum when you are summing different numbers?

    It is not to difficult to show that if you have any arithmetic sequence: an= a+ id, where a and d are fixed and i ranges from 1 to n, sums to n times the average of a1 and an: n*(a+ d+ (a+ nd))/2= n(a+ (d/2)(n+1)).

    In the case of 2+ 4+ 6+ ...+ 2n, a= 0 and d= 2. The sum is n(0+ (2/2)(n+1)= n(n+1)
    In the case of 1+ 3+ 5+ ...+ 2n+1, a= -1 and d= 2. The sum is n(-1+ (2/2)(n+1))= n2.
     
  6. Oct 16, 2008 #5
    Re: Recursion

    Thank you all for your help in explaining this, I appreciate it greatly.
     
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