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anonymous

  • 5 years ago

What is the nth term of the sequence 3, 11, 31, 69, 131?

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  1. anonymous
    • 5 years ago
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    Have you tried some sort of finite difference method?

  2. anonymous
    • 5 years ago
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    I had a look at it, and I'm thinking the nth term is a polynomial in n.

  3. anonymous
    • 5 years ago
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    If you set it up as: 3 11 31 69 131 8 20 38 62 12 18 24 6 6 0

  4. anonymous
    • 5 years ago
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    If you take the first line as some function of n, f(n) (i.e. zeroth derivative), then the fifth line will be the fourth derivative and you can start integrating back up. So\[\frac{d^4f(n)}{dn^4}=0 \rightarrow \frac{d^3f(n)}{dn^3}=c_1 \rightarrow \frac{d^2f(n)}{dn^2}=c_1n+c_2\]\[\rightarrow \frac{df(n)}{dn}=\frac{c_1n^2}{2}+c_2n+c_3 \]\[\rightarrow f(n)=\frac{c_1}{6}n^3+\frac{c_2}{2}n^2+c_3n+c_4\]

  5. anonymous
    • 5 years ago
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    You can then set up four linear equations with the first four terms and solve for the coefficients. I did this with a matrix calculator. When I tested for the 5th term, it spat out 130.29. The discrepancy probably due to error in rounding at each step.

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