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Sum of first n terms of a geometric series ? 1 + 2 + 2^2 +...+ 2^n-1 + 2^n watch

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    I typed it incorrectly in the title

    The Series is 1 + 2 + 2^2 +...+ 2^n-1 + 2^n

    How do I find the sum of the above series ?
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    (Original post by James222)
    I typed it incorrectly in the title

    The Series is 1 + 2 + 2^2 +...+ 2^n-1 + 2^n

    How do I find the sum of the above series ?
    This is a GP and so the standard formula applies, as your first term is 1.
    a=1, r=2 S_n = \frac{1(r^n-1) }{r-1}

    So, as an example-test, for the first 4 terms, n=4
    S_4 = \frac{1(2^4-1)}{2-1}=15 Which is 1+2+4+8.
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    (Original post by nerak99)
    This is a GP and so the standard formula applies, as your first term is 1.
    a=1, r=2 S_n = \frac{1(r^n-1) }{r-1}

    So, as an example-test, for the first 4 terms, n=4
    S_4 = \frac{1(2^4-1)}{2-1}=15 Which is 1+2+4+8.
    Thanks but my textbook has the following answer http://imgur.com/a/xke9G

    Any idea why they have r^n+1 rather than just r^n ?
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    Not really. If you put in t=1 into their formula for yt, you get y1=3 ?
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    (Original post by nerak99)
    Not really. If you put in t=1 into their formula for yt, you get y1=3 ?
    Perhaps they have a different formula?
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    There are n+1 terms in the sequence, Therfore \displaystyle S_{n+1} = \frac{a(1-r^{n+1})}{1-r}
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    On the face of it maths and chess is right. However In the scan they refer to 1 as the first term, not the zeroeth. Thus the nth term is 2^(n-1). Then in the formula for yt they seem to count from 0. I think it is the text that is not clear.

    To compound the confusion they then quote the formula that I used.

    The usually quoted formula is for the sum to n terms.

    However, please clear up my confusion.
 
 
 
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