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Math Help - Number sequences

  1. #1
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    Number sequences

    This has been bugging me for a while:

    If I have a number sequence for example, 1, 2, 4, 7, 11
    So its like adding 1 then 2 then 3 then 4 and so on..

    How would I write it in terms of n?

    Also,

    My teacher had a method which I dodnt write down in year 9 (am now in year 11) that you wrote a number sequence then drew an arrow down to the number sequence in the question and saw the relationship im wondering if anyone is familiar with that?

    Please reply
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  2. #2
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    Quote Originally Posted by Trin769 View Post
    This has been bugging me for a while:
    If I have a number sequence for example, 1, 2, 4, 7, 11
    Go to this site and simply type in the sequence.
    The On-Line Encyclopedia of Integer Sequences

    I do hate these problems because I think they serve no purpose.
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  3. #3
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    f(1) = 1, \ f(n)=\frac{n(n+1)}{2} +1 \mbox{ for } n >1
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  4. #4
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    Quote Originally Posted by VENI View Post

    f(1) = 1, \ f(n)=\frac{n(n+1)}{2} +1 \mbox{ for } n >1
    that's not quite right becasue it doesn't give us the second term, which is 2. the correct form is this: f(n)=\frac{n(n-1)}{2} + 1, \ n \geq 1.
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  5. #5
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    Smile check this out!!

    If u closely see the sequence......u can come up with this relation

    f(n) = f(n-1) + n-1 where f(1)=1 ---------(1)

    Follow thes steps....

    f(n) = f(n-1) + n-1 //Now substitute for f(n-1) using ----(1)
    = [f(n-2) + n-2] + n-1
    = f(n-2) + 2n -(1+2)
    = [f(n-3) + n-3)] + 2n -(1+2) //Now substitute for f(n-2) using ---(1)
    = f(n-3) +3n - (1 + 2 + 3)
    ....
    ....
    = f(n-k) + kn -(1+2+3 +.......k)

    Now we know the value of f(1)=1.....soo.....to make f(n-k) =f(1).....n-k=1 => k=n-1
    then the abv becmes....

    =f(1) + (n-1)n -(n-1)n/2
    =1 + n(n-1)/2

    Done!!
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  6. #6
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    Thanks so much for all the replies

    But it seems awfully complicated - I am doing GCSE maths at the moment.. :S

    Is there an easier way to work out a number sequence which isnt linear?
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  7. #7
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    Hello, Trin769!

    Is there an easier way to work out a number sequence which isn't linear?

    If we know that the formula is a quadratic, there is a sneaky way,
    . . but you may need some background.


    We begin with a basic quadratic function: . f(n) \:=\:\frac{n(n+1)}{2}

    This generates the Triangular Numbers: 1, 3, 6, 10, 15, . . .

    And it should be obvious why they are called "triangular."
    Code:
                                              o
                              o             o   o
                  o         o   o         o   o   o
          o     o   o     o   o   o     o   o   o   o
    
          1       3           6               10

    So we have the formula: .  T(n) \:=\:\frac{n(n+1)}{2}

    . . and we will modify it to fit the given sequence: 1, 2, 4, 7, 11, 16, 22, . . .


    For the first few values, compare our sequence a(n) to the triangular numbers T(n)

    . . \begin{array}{c||ccccccc}<br />
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline\hline<br />
a(n) & 1 & 2 & 4 & 7 & 11 & 16 & 22 \\<br />
T(n) & 1 & 3 & 6 & 10 & 15 & 21 & 28 \end{array}


    The values of T(n) do not match those of a(n).
    . . Some adjustments are necessary . . . let's examine them.

    . . \begin{array}{c||ccccccc}<br />
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline\hline<br />
a(n) & 1 & 2 & 4 & 7 & 11 & 16 & 22 \\<br />
T(n) & 1 & 3 & 6 & 10 & 15 & 21 & 28 \\ \hline<br />
{\color{red}\text{adj}} & {\color{red}\text{-}0} & {\color{red}\text{-}1} & {\color{red}\text{-}2} & {\color{red}\text{-}3} & {\color{red}\text{-}4 }& {\color{red}\text{-}5} & {\color{red}\text{-}6}<br />
 \end{array}


    We see that we must subtract a number one less than n.

    . . Hence: . a(n) \;=\;T(n) - (n-1)

    And we have: . a(n) \;=\;\frac{n(n+1)}{2} - (n-1)

    . . Therefore: . \boxed{a(n) \;=\;\frac{n^2-n+2}{2}}

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  8. #8
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    Thanks so much Soroban! I actually get that


    *victory dance*

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