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lilsis76

  • 2 years ago

How can I re-write the sum's suing Sigma Notation for: 2 - 2^2 + 2^3 -2^4 +2^5 I know the reoccuring number is going to be 2 which goes to the right of the sigma. on the bottom of sigma it would be 1, and top of sigma would be 5 right?

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  1. lilsis76
    • 2 years ago
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    |dw:1353786244962:dw|

  2. cnknd
    • 2 years ago
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    \[\sum_{n=0}^{4}2(-2)^n\] is one way of writing it.

  3. lilsis76
    • 2 years ago
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    oh. so that (-2), is that how it would go from negative to positive?

  4. cnknd
    • 2 years ago
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    expand what i wrote and check for yourself

  5. lilsis76
    • 2 years ago
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    okay: -2^0 = 0 *2 = 0 -2^1=-2*2=-4 -2^2=-4*2=-8 -2^3=-8*2=-16

  6. lilsis76
    • 2 years ago
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    @cnknd is this how it would go?

  7. cnknd
    • 2 years ago
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    ur doing the exponents wrong... this is basics...

  8. lilsis76
    • 2 years ago
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    |dw:1353786706417:dw|okay the first one would be>....< @cnknd

  9. cnknd
    • 2 years ago
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    in what i put down, the index n goes from 0 to 4... so it's 2(-2)^0 + 2(-2)^1 + 2(-2)^2 + 2(-2)^3 + 2(-2)^4 which is 2(1) + 2(-2) + 2(4) + 2(-8) + 2(16)

  10. lilsis76
    • 2 years ago
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    OH SHOOT! i just noticed i started at 1. i just now noticed it starts at 0

  11. lilsis76
    • 2 years ago
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    okay okay, i got it now. it makes more sense starting at zero haha thank you. Medal for you :)

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