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    Someone sent me a PM asking that I help with sequences... like
    1, 6, 18,40,75
    where you have to find the nth term. I went about my weird devised trial and error method I kind of fluked in GCSE, and found the second differnce to be 3, when I realised this was actually the second derivative of a function f(n), which gives the nth term in the series!

    so integrating,

    f''(n) = 3
    f(n) = (3n^2)/2 + Cn + D
    where C and D are constants of first and second integrations respectively/

    weird because the last time I thought about sequences was when I was halfway through GCSE, and facing a question I had to use SATs knowledge to tackle.

    then you can use trial and adjustment to find C and D.
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    Er, no.

    Let us assume you have the correct sequence. I.e f(N)=(3n^2)/2 + Cn + D

    Then f(1) = 3/2 + C + D = 1
    and f(2) = 6 + 2C + D = 6

    So C + D = -1/2 and 2C + D = 0

    So C = 1/2 and D = -1

    So f(N) = (3n^2)/2 + n/2 -1

    And f(3) = 13.5 + 1.5 - 1 = 14 which is not equal to 18.
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    So theyre not the same?
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    I dont think so, but I could be wrong, if it takes n sets of differences for the differences become constant, then the formula for each term will be a polynomial of degree n. I think its just coincidence that integrating a constant n times also produces a polynomial of degree n. I cant see any reason why the two should be related.
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    BTW, its the 3rd difference that is 3, not the second.
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    I just assumed that f'''(N) = 3 and using f(1), f(2) and f(3) I got f(N) = 1/2N^3 + 1/2N, which works for f(1), f(2) and f(3).

    I think your method simply generates a polynomial with the property that it evaluates to the chosen values used to find the formula. It isn't telling you anything special about the sequence.
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    we were told that if the second difference didn't change, then you halve the square of n and use that to get an approximation

    kind of made me think... second derivative is constant, second order polynomial
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    Given any sequence of numbers you can generate a polynomial which will evaluate to that sequence if you use N = 1, 2, 3 etc.

    So I can generate a polynomial for:

    A = {1, 6, 18, 40, 75, 2...
    B = {1, 6, 18, 40, 75, 20...
    C = {1, 6, 18, 40, 75, 80...
    D = {1, 6, 18, 40, 75, 10000000...
    E = {1, 6, 18, 40, 75, 5.43334433...
    F = {1, 6, 18, 40, 75, 0...

    So giving us S = {1, 6, 18, 40, 75} and asking to find the nth term is meaningless. That is why I am always annoyed why IQ tests ask you to generate the next term in the sequence, i.e {2, 3, 5, 7, n...

    What is n? It could be 9 i.e the next prime number or it could be 10, or it could be 434. There is no way to tell without more information. When these are given as GCSE questions they are normally limited to polynomials of order less than n^2, which are relatively easy to work out. But mathematically the numbers you have given do not uniquely determine a sequence.
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    Pentagonal pyramidal numbers: n^2 (n + 1)/2.

    http://www.research.att.com/~njas/sequences/

    If you know the formula is cubic (eg, from taking differences) then you can write it as

    A(n - 1)(n - 2)(n - 3)
    +
    B(n - 1)(n - 2)(n - 4)
    +
    C(n - 1)(n - 3)(n - 4)
    +
    D(n - 2)(n - 3)(n - 4),

    and then set n = 1, 2, 3, 4 to get the constants.
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    Ah yes, I made an error earlier.

    f(n)=1/2N^2 + 1/2N^3, and that actually works for all the given values. However I can still generate a polynomial for the sequence 1, 6, 18, 40, 75, 2 to demonstrate my point. It looks ugly, but it works:

    f(N) = 124 - (4274/15)*x +233x^2 - (262/3)*x^3 +(31/2)*x^4 - (31/40)*x^5
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    (Original post by JamesF)
    I dont think so, but I could be wrong, if it takes n sets of differences for the differences become constant, then the formula for each term will be a polynomial of degree n. I think its just coincidence that integrating a constant n times also produces a polynomial of degree n. I cant see any reason why the two should be related.
    Yes, this is right. It follows say from the fact that

    Sum[r=1->r=n] r^k

    is a polynomial of order k+1 in n. The first term is n^(k+1)/(k+1) but the other coefficients are complicated and involve things called Bernoulli numbers.

    I agree with other comments that it's a poor decision to give only five terms of a sequence where the expected answer is a cubic. After all through any five terms you can find a quartic.
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    Here's a sequence deliberately set up to prove the nonsense of these "what number comes next problems?"

    What's the missing number in the sequence:

    7, 9, 12, ?, 24, 36, 56, 90

    With some thought (and a knowledge of limits) you eventually twig that the answer is

    24 ln 2
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    But it could also be 7 .
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    1, 1, 1, 1, 1, ...

    Obviously 42.

    nth term = (41/120)n^5 - (41/8)n^4 + (697/24)n^3 - (615/8)n^2 + (5617/60)n - 40
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    (Original post by SsEe)
    1, 1, 1, 1, 1, ...

    Obviously 42.

    nth term = (41/120)n^5 - (41/8)n^4 + (697/24)n^3 - (615/8)n^2 + (5617/60)n - 40
    lol
    well shown
 
 
 
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