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Pologirl19

What is the sum of the geometric series

  • one year ago
  • one year ago

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  1. Pologirl19
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    \[\sum_{x=1}^{10}6(2)^x\]

    • one year ago
  2. satellite73
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    \[6\sum_{x=1}^{10}2^x\] is a start

    • one year ago
  3. satellite73
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    \[\sum_{x=1}^{10}2^x=2^{11}-2\]

    • one year ago
  4. Pologirl19
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    where did you get 2^11 - 2

    • one year ago
  5. Pologirl19
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    @satellite73

    • one year ago
  6. General_Gustav
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    The sum of this series involves factoring out the 6 into the front of the equation, like satellite73 said. I'm sure there is a shortcut to this, but i would write out each \[2^{x}\] term with n beginning with 1 and ending with 11 and then add them all together. Probably not the answer that you're looking for, but at least its a start.

    • one year ago
  7. tkhunny
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    \(\sum\limits_{x=1}^{10} 2^{x} = 2 + 2^{2} + 2^{3} + ... + 2^{10}\) Using the usual technique of creating a Sum and multiplying by the common factor, we get \(\dfrac{2 - 2^{11}}{1 - 2} = 2^{11} - 1\)

    • one year ago
  8. tkhunny
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    * - 2 Typo, sorry.

    • one year ago
  9. General_Gustav
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    ^This is the shortcut I was mentioning. :D

    • one year ago
  10. satellite73
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    in general \[\sum_{k=1}^n r^k = \frac{r^{n+1}-r}{r-1}\]

    • one year ago
  11. satellite73
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    in your case \(r=2\) so you get \[\frac{2^{11}-2}{2-1}=2^{11}-2\]

    • one year ago
  12. Pologirl19
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    That is none of my answers the equation for it is \[Sn=a1(\frac{ 1- r^n }{ 1 - r })\]

    • one year ago
  13. Pologirl19
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    @satellite73

    • one year ago
  14. tkhunny
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    @Pologirl19 We ran into this problem the other day. 1) The ability to create the formula is worth far more than a list of formulas you don't understand. 2) If you had the formula all along, why didn't you use it? 3) A little flexibility and thought will produce n = 10, a1 = 2, and r = 2 and this is an equivalent expression. Please brush up on your algebra.

    • one year ago
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