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Thread: Formula to reflect sequence of numbers

  1. #1
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    Formula to reflect sequence of numbers

    Is there a formula to reflect this sequence of numbers so that, for instance I could show the calculations for 8^3 without having to do the preceding calculations?
    1^3=1 1+(6x0)=1+0=1
    2^3=8 1+(6x1)=7+1=8
    3^3=27 7+(6x2)=19+8=27
    4^3=64 19+(6x3)=37+27=64
    5^3=125 37+(6x4)=61+64=125
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  2. #2
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    Re: Formula to reflect sequence of numbers

    Quote Originally Posted by tridoddytri View Post
    Is there a formula to reflect this sequence of numbers so that, for instance I could show the calculations for 8^3 without having to do the preceding calculations?
    $(N+1)^3=N^3+3N^2+3N+1=N^3+3N(N+1)+1$

    $\mathcal{a}_{1}=1$, if $N\ge 1~~\mathcal{a}_{N+1}=N^3+3N(N+1)+1$
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  3. #3
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    Re: Formula to reflect sequence of numbers

    Isn't 8 \times 8 \times 8 easier to calculate?

    We could try to formalise your method. We seem to have two sequences \{a_n\} the result of the first part of the calculation and \{b_n\} for which b_n = n^3, but we won't use this fact.

    Your calculation appears to be as follows:
    a_{n+1} = a_{n} + 6n \qquad b_n = a_n + b_{n-1}
    a_0=1 \qquad b_0 = 0

    The first equation is easily solved
    \begin{aligned} a_{n+1} - a_{n} = 6n \implies a_{n+2} - a_{n+1} &= 6(n+1) = 6n + 6 \\ a_{n+2} - a_{n+1} &= a_{n+1} - a{n} + 6 \\ a_{n+2} - 2a_{n+1} + a_{n} &= 6 \\ \implies a_{n+3} - 2a_{n+2} + a_{n+1} &= 6 \\ \text{and so} \quad a_{n+3} - 2a_{n+2} + a_{n+1} &= a_{n+2} - 2a_{n+1} + a_{n} \\ \text{and finally} \quad a_{n+3} - 3a_{n+2} + 3a_{n+1} - a_n &= 0 \end{aligned}

    This has characteristic equation (r-1)^3 = 0 so our solution will be
    a_n = (An^2 + Bn + C)1^n = An^2 + Bn + C

    We know that a_0 = 1, a_1 = 1 and a_2 = 7 so we deduce that C=1, A+B+1=1 and 4A + 2B +1 = 7 which result in A = 3 and B = -3.

    Thus we have
    a_n = 3n^2 - 3n + 1

    We can now put this into our equation for b_{n+1} getting (after a bit of rearranging)
    b_{n+1} - b_n = 3n^2 - 3n + 1

    We can solve this using the same method.
     \begin{aligned} b_{n+2} - b_{n+1} &= 3(n+1)^2 - 3(n+1) + 1 = 3n^2 + 3n  + 1 \\ b_{n+2} - 2b_{n+1} +b_n &= 6n \\ \implies b_{n+3} - 2b_{n+2} +b_{n+1} &= 6(n+1) + 1 = 6n + 1 \\ b_{n+3} - 3b_{n+2} + 3b_{n+1} - b_{n} = 1 \\ \implies b_{n+4} - 3b_{n+3} + 3b_{n+2} - b_{n+1} &= 1 \\ \text{and so} \quad b_{n+4} - 4b_{n+3} + 6b_{n+2} - 4b_{n+1} + b_{n} &= 0\end{aligned}
    And the characteristic equation (r-1)^4 = 0 gives us a solution b_n = (pn^3 + qn^2 + rn + s)1^n = pn^3 + qn^2 + rn + s. But we know that b_0 = 0, b_1 = 1, b_2 = 8, b_3 = 27 which results in q=r=s=0 and p=1. And thus
    b_n = n^3
    Are you surprised?
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  4. #4
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    Re: Formula to reflect sequence of numbers

    Let's use n^3

    You apparently want to see a "breakdown" of the calculation...

    Li'l computer program:
    ...........................................
    INPUT n [n = 5]

    a = 3(n^2 - 3n + 2) + 1
    b = 6(n - 1)
    c = (n - 1)^3

    PRINT a, b, c, a+b+c
    ...........................................

    OUTPUT:
    37, 24, 64, 125 : that's what you show for 5^3

    If n = 8:
    127, 42, 343, 512

    Kinda silly, if you ask me...
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