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Math Help - Sequences : Question

  1. #1
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    Sequences : Question

    Hi All,
    I was wondering if you could help me with the below...

    . If the sequence x(1), x(2), x(3), , x(n), is such that x(1) = 3 and x(n+1) = 2x(n) 1 for n ≥ 1, then x(20) x(19) =

    A. 2^19
    B. 2^20
    C. 2^21
    D. (2^20) - 1
    E. (2^21) - 1

    I'm drawing a complete blank. x(n+1) = 2x(n) - 1 would ellude to x(20) being 2x(value of 19) -1. = 2x(19) - 1 = 38x - 1 which seems completely off? Do I have to substitute all the way from x(1) to x(19) to determine the value?

    Thanks in Advance.
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  2. #2
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    2x(19) - 1 = 38x - 1 which seems completely off?
    Yes, this is off because x is a function from natural numbers to natural numbers; it is not a number. Thus, x(19) is a number, but 2x and 38x don't make sense.

    I'll write the argument of x as a subscript: x_n instead of x(n). By definition, x_{20}-x_{19} = 2x_{19}-1-x_{19}=x_{19}-1. After that, the first idea that came to mind was to guess the general formula for x_n by calculating the first several values of x_n and, possibly, to prove this general formula by induction.

    \begin{array}{c|c|c|c|c|c|c|c}<br />
n & 1 & 2 & 3 & 4 & 5 & 6 & 7\\<br />
\hline<br />
x_n & 3 & 5 & 9 & 17 & 33 & 65 & 129<br />
\end{array}
    Can you guess the general formula for x_n?
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  3. #3
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    Quote Originally Posted by emakarov View Post
    Yes, this is off because x is a function from natural numbers to natural numbers; it is not a number. Thus, x(19) is a number, but 2x and 38x don't make sense.

    I'll write the argument of x as a subscript: x_n instead of x(n). By definition, x_{20}-x_{19} = 2x_{19}-1-x_{19}=x_{19}-1. After that, the first idea that came to mind was to guess the general formula for x_n by calculating the first several values of x_n and, possibly, to prove this general formula by induction.

    \begin{array}{c|c|c|c|c|c|c|c}<br />
n & 1 & 2 & 3 & 4 & 5 & 6 & 7\\<br />
\hline<br />
x_n & 3 & 5 & 9 & 17 & 33 & 65 & 129<br />
\end{array}
    Can you guess the general formula for x_n?
    ekkk Is it ....
    X_n = 2(X_{n-1} - X_{n-2}) + X_{n-1} ?

    Could you elaborate on the substitution that resulted in x_{20}-x_{19} = 2x_{19}-1-x_{19}=x_{19}-1. I'm not sure where the  -x_{19} came from?
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  5. #5
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    Quote Originally Posted by dumluck View Post
    ekkk Is it ....
    X_n = 2(X_{n-1} - X_{n-2}) + X_{n-1} ?
    By "general formula" I meant "a closed-form solution: a non-recursive function of n" (Wikipedia). Your expression for x_n is still recursive because it refers (recurs) to x_{n-1}.

    Could you elaborate on the substitution that resulted in x_{20}-x_{19} = 2x_{19}-1-x_{19}=x_{19}-1.
    By definition, x_{n+1}=2x_{n}-1. Substituting n = 19, we get x_{20}=2x_{19}-1. Subtracting x_{19} from both sides, we get x_{20}-x_{19}=2x_{19}-1-x_{19}.
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  6. #6
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    Quote Originally Posted by emakarov View Post
    By "general formula" I meant "a closed-form solution: a non-recursive function of n" (Wikipedia). Your expression for x_n is still recursive because it refers (recurs) to x_{n-1}.

    By definition, x_{n+1}=2x_{n}-1. Substituting n = 19, we get x_{20}=2x_{19}-1. Subtracting x_{19} from both sides, we get x_{20}-x_{19}=2x_{19}-1-x_{19}.
    Ah yes.. thanks. Would this not be it then

    x_{n} = 2x_{n} - 1
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  7. #7
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    Quote Originally Posted by dumluck View Post
    Would this not be it then

    x_{n+1} = 2x_{n} - 1
    This is the definition, not the solution. As Plato's link says, x_n=2^n+1.
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  8. #8
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    Hello, dumluck!

    emakarov gave you an excellent hint.
    . . But you didn't catch it, did you?


    \begin{array}{c|c|c|c|c|c|c|c}<br />
n & 1 & 2 & 3 & 4 & 5 & 6 & 7\\<br />
\hline<br />
x_n & 3 & 5 & 9 & 17 & 33 & 65 & 129<br />
\end{array}

    Can you guess the general formula for \,x_n?

    Each number is one more than a power of 2.

    . . \begin{array}{ccc}<br />
3 &=& 2^1 + 1 \\<br />
5 &=& 2^2 + 1 \\<br />
9 &=& 2^3 + 1 \\<br />
17 &=& 2^4 + 1 \\<br />
33 &=& 2^5 + 1 \\<br />
\vdots && \vdots \end{array}

    Hence: . x(n) \:=\:2^n+1


    Therefore: . x(20) - x(19) \;=\;(2^{20}-1) - (2^{19}-1)

    . . . . . . . . . . =\;2^{20} - 2^{19} \;=\;2^{19}(2 - 1) \;=\;2^{19} . answer A

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  9. #9
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    ahh yes I see. Although the level of hint I would need to get that would be ... It's something to the power of 2 . So all I had to do was transform the definition to the solution for Xn? moving the x to the other side. Thanks guys, much appreciated. Learnt a lot there. I love this forum.
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  10. #10
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    Quote Originally Posted by dumluck View Post
    Hi All,
    I was wondering if you could help me with the below...

    . If the sequence x(1), x(2), x(3), , x(n), is such that x(1) = 3 and x(n+1) = 2x(n) 1 for n ≥ 1, then x(20) x(19) =

    A. 2^19
    B. 2^20
    C. 2^21
    D. (2^20) - 1
    E. (2^21) - 1

    I'm drawing a complete blank. x(n+1) = 2x(n) - 1 would ellude to x(20) being 2x(value of 19) -1. = 2x(19) - 1 = 38x - 1 which seems completely off? Do I have to substitute all the way from x(1) to x(19) to determine the value?

    Thanks in Advance.

    The possible solutions given are all in terms of "powers of 2".

    Hence the clue is in the question itself!
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