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Math Help - Sequences - nth Term

  1. #1
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    Exclamation Sequences - nth Term

    Hi. Please can you help me? I have three Maths tests next week, and we have a revision sheets which says lots of stuff, and the last one I can't do! It says nth term. I can do nth term all right, but I was wondering if there is an equation for a sequence if each time the gap changes, eg;
    8, 11, 15, 20, 26
    +3 +4 +5 +6

    __n +/- ___.

    Thanks. K
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by kate View Post
    Hi. Please can you help me? I have three Maths tests next week, and we have a revision sheets which says lots of stuff, and the last one I can't do! It says nth term. I can do nth term all right, but I was wondering if there is an equation for a sequence if each time the gap changes, eg;
    8, 11, 15, 20, 26
    +3 +4 +5 +6

    __n +/- ___.

    Thanks. K
    a_1=8

    a_2=8+(2+1)

    a_3=8+(2+1)+(2+2)=8+2\times2+(1+2)

    a_4=8+(2+1)+(2+2)+(2+3)=8+3\times2+(1+2+3)

    :
    :

    a_n=8+(n-1)\times 2+\sum_{r=1}^{n-1}r

    Where the last term is the sum of the first n-1 integers so:

    a_n = 8+2(n-1)+\frac{(n-1)n}{2}

    RonL
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  3. #3
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    Hello, Kate!

    Here's another approach . . .


    Take the difference of consecutive terms.
    Then take the differences of the differences, and so on
    . . until you get a row of constants.

    . . \begin{array}{cccccccccc}\text{Sequence:} & 8 & & 11 & & 15& & 20 & & 26 \\<br />
\text{1st di{f}ferece:} & & 3 & & 4 & & 5 & & 6 &\\<br />
\text{2nd dif{f}erence:} & & & 1 & & 1 & & 1 & & \end{array}

    The row of constants appeared in the second differences.
    . . Hence, the generating function is of the second degree, a quadratic.

    There are a number of methods for determining the generating function.
    . . Here's one of them . . .


    We know that the function has the form: . f(n) \:=\:an^2 + bn + c

    Using the first three values of the sequence, we have:
    . . \begin{array}{ccccccc}f(1) = 8: & a + b + c & = & 8 & [1]\\<br />
f(2) = 11: & 4a + 2b + c & = & 11 & [2]\\<br />
f(3) = 15: & 9a + 3b + c & = & 15& [3]\end{array}


    Then solve the system of equations . . .

    Subtract [1] from [2]: . 3a + b \:=\:3\;\;\;[4]
    Subtract [2] from [3]: . 5a + b \:=\:4\;\;\;[5]

    Subtract [4] from [5]: . 2a = 1\quad\Rightarrow\quad\boxed{a = \frac{1}{2}}

    Substitute into [4]: . 3\left(\frac{1}{2}\right) + b\:=\:3\quad\Rightarrow\quad\boxed{b = \frac{3}{2}}

    Substitute into [1]: . \frac{1}{2} + \frac{3}{2} + c \:=\:8\quad\Rightarrow\quad\boxed{c = 6}


    Therefore: . f(n) \;=\;\frac{1}{2}n^2 + \frac{3}{2}n + 6\quad\Rightarrow\quad \boxed{f(n)\;=\;\frac{1}{2}(n^2 + 3n + 12)}

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