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Math Help - Sequence problem

  1. #1
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    Sequence problem

    Determine the value for n for which

    (2^1)(2^2)(2^3)(2^4)...(2^n)=(2^210)

    I don't know what do here to be honest. Any help would be greatly appreciated, thanks!

    EDIT: it's 2 to the power of n = 2 to the power of 210, if that's unclear.
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  2. #2
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    Hello, Dovud!

    Determine the value for n for which:

    . . (2^1)(2^2)(2^3)(2^4)\:\cdots\:(2^n)\;=\;2^{210}

    We have: . 2^{(1+2+3+\cdots+n)} \;=\;2^{210}

    . . Hence: . 1 + 2 + 3 + \cdots + n \;=\;210


    The sum of the first n natural numbers is given by: . \frac{n(n+1)}{2}

    . . So we have: . \frac{n(n+1)}{2} \:=\:210 \quad\Rightarrow\quad n^2 + n - 420 \:=\:0

    . . Factor: . (n-20)(n+21) \:=\:0 \quad\Rightarrow\quad n \;=\;20,\;{\color{red}\rlap{/////}}-21


    Therefore: . {\color{blue}n \;=\;20}

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  3. #3
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    Thank you!
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  4. #4
    Super Member Bacterius's Avatar
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    EDIT: it's 2 to the power of n = 2 to the power of 210, if that's unclear.
    To write more complicated exponents into LaTeX, insert them into braces :

    2^{210} \Rightarrow 2^{210}
    2^{3x + \sqrt{4}} \Rightarrow 2^{3x + \sqrt{4}}

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