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  • 5 years ago

Find the Geometric sum

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  1. anonymous
    • 5 years ago
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    \[\sum_{k=1}^{\infty} 1/5^k\]

  2. anonymous
    • 5 years ago
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    Now you have to do a little bit of manipulation to get this in the appropriate form. I'm assuming you mean \[1/(5^{k})\] right? If so, then you have to use the geometric series form: \[\sum_{k=1}^{\infty}ar^{n-1}\] However, we are missing the n-1 so we'll have to do some algebra trickery to get it in this form. An important think to note is that 1 to any power will always remain 1 so the series can be changed to the following: \[\sum_{k=1}^{\infty}1^{k-1}/5^{k}\]This may not seem right, but if you think about it, this won't change the sum because each term in the sum will have a 1 in the numerator. Now for the 5^k, we can do a little trick by raising 5^(k-1+1)=5^k so, we can change the series again to: \[\sum_{k=1}^{\infty}1^{k-1}/5^{k-1+1}\]Now when you have the same base multiplied together with different powers, you can add the powers together, i.e.\[x^{a}x^{b}=x^{a+b}\]So we can pull out a 5 from the denominator in the sum above to get the new sum: \[\sum_{k=1}^{\infty}1^{k-1}/(5^{k-1}*5)\] Rewriting to make it cleaner:\[\sum_{k=1}^{\infty}(1/5)(1/5)^{k-1}=(1/5)/(1-1/5)=(1/5)/(4/5)=1/4\]

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