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gjhfdfg

  • one year ago

Finding the indicated sum ,

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  1. gjhfdfg
    • one year ago
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  2. whpalmer4
    • one year ago
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    This one can be solved without any cleverness. \[\sum_{k=1}^{4}(-1)^k(k+11) =\]\[(-1)^1(1+11)+(-1)^2)(2+11)+(-1)^3)(3+11)+(-1)^4(4+11) =\]

  3. whpalmer4
    • one year ago
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    Now why did I stick in those unwanted )'s? \[(-1)^1(1+11)+(-1)^2(2+11)+(-1)^3(3+11)+(-1)^4(4+11) =\] For odd k, \((-1)^k = -1\) For even k, \((-1)^k = 1\) -(1+11)+(2+11)-(3+11)+(4+11) = all the alternating of signs cancels out the 11s, so it's just -1 + 2 - 3 + 4 = 2

  4. gjhfdfg
    • one year ago
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    How did you get the -1+2-3+4?

  5. gjhfdfg
    • one year ago
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    When I solved it I didnt get none of that, maybe I did it wrong?

  6. whpalmer4
    • one year ago
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    \[-(1+11)+(2+11)-(3+11)+(4+11) = -1 -11 + 2 + 11 - 3 - 11 + 4 + 11\] The alternating -11 + 11 sum to 0, leaving -1 + 2 - 3 + 4

  7. gjhfdfg
    • one year ago
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    Ah okay, I get it

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