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    They're not equivalent.

    \displaystyle\sum_{k=1}^\infty \left(\frac{1}{2}\right)^k  \left(\frac{11}{12}\right)^{k-1} = \sum_{k=1}^\infty \left(\frac{1}{2}\right) \left(\frac{1}{2}\right)^{k-1}  \left(\frac{11}{12}\right)^{k-1} =  \sum_{k=1}^\infty \left(\frac{1}{2}\right)   \left(\frac{11}{24}\right)^{k-1} =  \frac{1}{2}  \sum_{k=1}^\infty  \left(\frac{11}{24}\right)^{k-1}
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    (Original post by ttoby)
    They're not equivalent.

    \displaystyle\sum_{k=1}^\infty \left(\frac{1}{2}\right)^k  \left(\frac{11}{12}\right)^{k-1} = \sum_{k=1}^\infty \left(\frac{1}{2}\right) \left(\frac{1}{2}\right)^{k-1}  \left(\frac{11}{12}\right)^{k-1} =  \sum_{k=1}^\infty \left(\frac{1}{2}\right)   \left(\frac{11}{24}\right)^{k-1} =  \frac{1}{2}  \sum_{k=1}^\infty  \left(\frac{11}{24}\right)^{k-1}
    Thanks for your reply! But according to the answer I have here they are equivalent, I'm not sure how they do it though. I don't think there is a mistake in the answer since its a mark scheme for an exam.
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    I agree with ttoby. I'd suggest posting a photo of the actual question.
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    (Original post by ghostwalker)
    I agree with ttoby. I'd suggest posting a photo of the actual question.
    Here's the question:

    I got up to the summation from k=1 to infinity of (1/2)^k (11/12)^(k-1)
    EDIT: Got rid of the question, mind if I pm you the question instead? Just found out I can't make the question publicly available...
    EDIT2: Nevermind got it sorted now, but have another question about this! How do I simplify the summation from k=1 to infinity of (11/14)^k-1?
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    (Original post by JBKProductions)
    EDIT2: Nevermind got it sorted now, but have another question about this! How do I simplify the summation from k=1 to infinity of (11/14)^k-1?
    Setting aside the factor of 1/2 you have a geometric progression. First term is? Common ratio is? Apply forumla a/(1-r) = result. Then divide by 2.

    Edit: I see the 1/2 has gone so ignore the bit about 1/2 and divide by 2
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    (Original post by JBKProductions)
    Here's the question:

    I got up to the summation from k=1 to infinity of (1/2)^k (11/12)^(k-1)
    EDIT: Got rid of the question, mind if I pm you the question instead? Just found out I can't make the question publicly available...
    EDIT2: Nevermind got it sorted now, but have another question about this! How do I simplify the summation from k=1 to infinity of (11/14)^k-1?
    It's the sum of an infinite geometric series. 1 + (11/14) + (11/14)^2 + ... = \dfrac{1}{1-\frac{11}{14}}
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    (Original post by ghostwalker)
    Setting aside the factor of 1/2 you have a geometric progression. First term is? Common ratio is? Apply forumla a/(1-r) = result. Then divide by 2.
    Thank you!
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    (Original post by ttoby)
    It's the sum of an infinite geometric series. 1 + (11/14) + (11/14)^2 + ... = \dfrac{1}{1-\frac{11}{14}}
    Thank you too!
 
 
 
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