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    3+6+12+......384
    Find the sum
    I've worked out the ratio is 2, what else do I do
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    (Original post by Lucofthewoods)
    3+6+12+......384
    Find the sum
    I've worked out the ratio is 2, what else do I do
    Yes, add them all up now.

    3 + 6 + 12 + 24 + 48
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    (Original post by Naruke)
    Yes, add them all up now.

    3 + 6 + 12 + 24 + 48
    is there not a specific formula?
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    (Original post by Lucofthewoods)
    is there not a specific formula?
    Yes, it's a shortcut. If you can't perform the shortcut add them all up manually.
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    (Original post by Naruke)
    Yes, it's a shortcut. If you can't perform the shortcut add them all up manually.
    Well what is the formula?
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    S_n=\frac{a(r^n-1)}{r-1}



a=3

r=2

    Find n by substituting 384 into the equation for a geometric series

    a_n=a(r)^{n-1}
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    (Original post by memeeee)
    S_n=\frac{a(r^n-1)}{r-1}



a=3

r=2

    Find n by substituting 384 into the equation for a geometric series

    a_n=a(r)^{n-1}
    Can you hint more at what to do? I'm so confused
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    (Original post by Lucofthewoods)
    Can you hint more at what to do? I'm so confused
    The sum is given by the formula she mentioned. The sum of the first n terms of a geometric sequence with common ratio r and first term a is given by S_n=\frac{a(1-r^n)}{1-r}

    You found r=2 which is correct, and you know that a=3

    The only thing left to do is find n and to do that you need to use the formula for the n^{th} term of a geometric sequence which is u_n=ar^{n-1}. So you would have 384=3\cdot 2^{n-1} because 384 is the last term, and solve for n to see up to which term of the sequence you are summing up.

    Then just substitute a, r, n into the summation formula.
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    (Original post by RDKGames)
    The sum is given by the formula she mentioned. The sum of the first n terms of a geometric sequence with common ratio r and first term a is given by S_n=\frac{a(1-r^n)}{1-r}

    You found r=2 which is correct, and you know that a=3

    The only thing left to do is find n and to do that you need to use the formula for the n^{th} term of a geometric sequence which is u_n=ar^{n-1}. So you would have 384=3\cdot 2^{n-1} because 384 is the last term, and solve for n to see up to which term of the sequence you are summing up.

    Then just substitute a, r, n into the summation formula.
    Thank you
 
 
 
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