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anonymous

  • 5 years ago

write a formula that produces the given terms of the sequence a_1= 1/8, a_2= -1/4, a_3= 1/2, a_4=-1, a_5=2

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  1. amistre64
    • 5 years ago
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    Put everything in the same pants...like denominators: +1/8 -2/8 +4/8 -8/8 +16/8 .... does this help?

  2. amistre64
    • 5 years ago
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    y = -2(-x)/8 ? nope.....

  3. anonymous
    • 5 years ago
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    umm...it needs to be written in "n" form lol i think....its an infinite sequence i know that

  4. amistre64
    • 5 years ago
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    yeah.... I need to learn them better. But, I can write a program that will spit out the numbers in that sequence :)

  5. anonymous
    • 5 years ago
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    ok, that'd be great :)

  6. amistre64
    • 5 years ago
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    ack!!.... maybe I cant

  7. anonymous
    • 5 years ago
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    \[a_n = (-1)^n*2^n*1/8\] I felt SO relieved after I got this. :P Btw, it starts at n=0.

  8. anonymous
    • 5 years ago
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    i thought it started at 1

  9. anonymous
    • 5 years ago
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    i get it now :)

  10. anonymous
    • 5 years ago
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    Sorry, if it starts with a_1 then it should be:\[a_n = (-1)^{n-1}*2^{n-1}*1/8\]

  11. amistre64
    • 5 years ago
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    Are we allowed to use trig? I had considered 2^n cos(pi(n-1)) / 8

  12. anonymous
    • 5 years ago
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    That doesn't quite work, the first term gives you something that isn't 1/8, but yours is very creative...

  13. amistre64
    • 5 years ago
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    granted, but if the sequence starts at a_1 it could work...maybe....with a little luck :) and thanx!

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