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Math Help - Quickie #16

  1. #1
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    Quickie #16


    Express the sum of the first n terms of this sequence:

    1,\;(1+2),\;(1+2+2^2),\;(1+2+2^2+2^3),\;\cdots,\;(  1 + 2 + 2^2 + \cdots + 2^{n-1})

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  2. #2
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    The sum of the first n terms is 2^{n+1}-(n+2)

    For instance, sum of first term, n=1: 2^{2}-3=1

    Sum of first 2 terms: 2^{3}-4=4

    Sum of first 5 terms: 2^{6}-7=57

    The terms are 1, 3, 7, 15, 31, 63,....,2^{n-1}
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  3. #3
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    Quote Originally Posted by galactus View Post
    The sum of the first n terms is 2^{n+1}-(n+2)

    For instance, sum of first term, n=1: 2^{2}-3=1

    Sum of first 2 terms: 2^{3}-4=4

    Sum of first 5 terms: 2^{6}-7=57

    The terms are 1, 3, 7, 15, 31, 63,....,2^{n-1}
    Are you sure?

    For when n=3, ans=7

    2^{3-1}=2^2=4??

    Don't you mean...

    1,3,7,15,31,63,....,2^n-1



    *Edit: Oh! Sorry...you got it right at the first sentance.. >_<
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  4. #4
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    Good work, Galactus!


    Here's my book's solution . . .

    The k^{th} term is: . (1 + 2 + 2^2 + \cdots + 2^{k-1}) \:=\:2^k - 1

    The sum is: . S_n\;=\;(2 - 1) + (2^2-1) + (2^3-1) + (2^4 - 1) + \cdots + (2^n-1)

    . . S_n\;=\;\underbrace{(2 + 2^2 + 2^3 + \cdots + 2^n)}_{\text{geometric series}} - \underbrace{(1 + 1 + 1 + \cdots + 1)}_{n\text{ terms}}

    . . S_n\;=\;2\,\frac{1-2^n}{1 - 2} - n\quad\Rightarrow\quad \boxed{S_n\;=\;2^{n+1} - 2 - n}

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