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Math Help - Formula for N

  1. #1
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    Formula for N

    Hiya

    In the number pathern 1.3.6.10.15.21.34
    I found the recursive formula an+1-an=n+1
    now i need to find a formula for n.
    shouldent it be somthing like an+1=n ?
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  2. #2
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    Quote Originally Posted by JBswe View Post
    Hiya

    In the number pathern 1.3.6.10.15.21.34
    I found the recursive formula an+1-an=n+1
    now i need to find a formula for n.
    shouldent it be somthing like an+1=n ?
    why wouldn't it be n = a_{n + 1} - a_n - 1?

    because n is actually not equal to a_{n + 1}, you can just look at the numbers and see that
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  3. #3
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    1, 3, 6, 10, 15, 21, 28, 36........

    Is an easy sequence.

    I take it you want to find an expression for the nth term?

    As a clue notice..
    1=1x2/2
    3=2x3/2
    6=3x4/2
    10=4x5/2
    etc.....

    The one you posted 1,3,6,10,15,21,34 is more difficult.
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  4. #4
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    Quote Originally Posted by a tutor View Post
    1, 3, 6, 10, 15, 21, 28, 36........

    Is an easy sequence.

    [snip]

    The one you posted 1,3,6,10,15,21,34 is more difficult.
    Well spotted. At this thread (which I assume post #1 is following up from) that's the sequence I had in mind too ..... 1,3,6,10,15,21,34 is more difficult and the formula I gave, a_{n+1} - a_n = n+1, clearly fails to give a_7 = 34 ....
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  5. #5
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    am sorry i messed up it is 1, 3, 6, 10, 15, 21, 28, not 34 as the 7th number
    Last edited by JBswe; January 15th 2008 at 09:57 AM.
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  6. #6
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    ok i did some recounting and the formula for n i think is an=n*(n-1/2) is this right?
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  7. #7
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    Quote Originally Posted by JBswe View Post
    ok i did some recounting and the formula for n i think is an=n*(n-1/2) is this right?

    No. Even if you meant n(n-1)/2

    This gives 0,1,3,6,10....
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  8. #8
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    Quote Originally Posted by a tutor View Post
    1, 3, 6, 10, 15, 21, 28, 36........

    Is an easy sequence.

    I take it you want to find an expression for the nth term?

    As a clue notice..
    1=1x2/2
    3=2x3/2
    6=3x4/2
    10=4x5/2
    etc.....

    The one you posted 1,3,6,10,15,21,34 is more difficult.
    A further clue is needed perhaps ....? Here's a massive one -

    1=1x(1+1)/2
    3=2x(2+1)/2
    6=3x(3+1)/2
    10=4x(4+1)/2
    etc.....

    The rule should come as no surprise since a_n is the sum of the first n positive integers ......

    btw 1 (as I suggested in the other thread) you could always look into the general method of solving a recurrence relation of the form a_{n+1} - a_n = \, \text{polynomial function of } n.

    btw 2 it's been said that mathematics is the study of patterns. And what better way of developing this than questions like this one ......
    Last edited by mr fantastic; January 15th 2008 at 01:00 PM. Reason: Tidied the latex
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  9. #9
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    ok think i got it

    n=n(n+1)/2
    and the recursive formula is
    an+1-an=n+1
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  10. #10
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    Quote Originally Posted by JBswe View Post
    ok think i got it

    a_n = n(n+1)/2 Mr F correction of typo(?)
    and the recursive formula is
    an+1-an=n+1
    Well you got there in the end. Thanks all round, I hope.
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  11. #11
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    Ya now that i know i got it down a big thanks to all of you who help me out, Thank the gods you are here
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