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Math Help - Determining the formula of this sequence?

  1. #1
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    Determining the formula of this sequence?

    I was given the sequence
    0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, ...
    and I'm looking for a method to determine an explicit formula for this.
    I've tried doing difference columns, and I always end up with 2^n being the 2nd difference.

    Anyone know a way to figure this out?
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  2. #2
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    Quote Originally Posted by paupsers View Post
    I was given the sequence
    0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, ...
    and I'm looking for a method to determine an explicit formula for this.
    I've tried doing difference columns, and I always end up with 2^n being the 2nd difference.

    Anyone know a way to figure this out?
    Good idea to look at differences. So you found:

    (u_{n+2}-u_{n+1})-(u_{n+1}-u_n)=2^n,

    assuming the first term is u_1=0.

    If you sum the above equalities for n ranging from 1 to n, you get a telescoping sum on the left, and a geometric sum on the right, which leads to:

    (u_{n+2}-u_{n+1})-(u_2-u_1)=2+2^2+\cdots +2^n = 2^{n+1}-2,

    hence

    u_{n+2}-u_{n+1}=2^{n+1}-1.

    Summing again this latter equality we have again a telescoping sum on the left and a geometric sum (minus constant terms) on the right, and we get:

    u_{n+2}-u_2=(2^2-1)+\cdots+(2^{n+1}-1),

    hence (change of variable)

    u_n = (2^2-1)+\cdots +(2^{n-1}-1)+1

    and finally u_n=2^n-n-1. Same method works in many situations.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by paupsers View Post
    I was given the sequence
    0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, ...
    and I'm looking for a method to determine an explicit formula for this.
    I've tried doing difference columns, and I always end up with 2^n being the 2nd difference.

    Anyone know a way to figure this out?
    The Encyclopaedia of Integer Sequences gives: 2^n-n-1

    CB
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  4. #4
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    Hello, paupsers!

    Given the sequence: . 0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, \hdots

    I'm looking for a method to determine an explicit formula for this.
    I've tried doing difference columns, and I always end up with 2^n being the 2nd difference.
    This is true . . . and should give you a big hint.
    Consider the general form: . f(n) = 2^n

    How does it compare with the given sequence?

    \begin{array}{c|cccccccccc}\hline<br />
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline<br />
\text{Sequence:} & 0 & 1 & 4 & 11 & 26 & 57 & 120 & 247 & 502 & 1013 \\ \hline<br />
f(n) = 2^n & 2 & 4 & 8 & 16 & 32 & 64 & 128 & 256 & 512 & 1024 \\ \hline<br />
\text{Error:} & +2 & +3 & +4 & +5 & +6 & +7 & +8 & +9 & +10 & +11\\ \hline<br />
\end{array}


    If we use f(n) = 2^n, each term is too large by (n+1)

    Therefore, the explicit formula is: . F(n) \;=\;2^n - (n+1)

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