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anonymous

  • 4 years ago

prove sigma n^2=n(n+1)(2n+1)/6

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  1. anonymous
    • 4 years ago
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    i know!

  2. anonymous
    • 4 years ago
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    that is \[\sum_{1}^{n}\]n^2

  3. anonymous
    • 4 years ago
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    then tell?

  4. anonymous
    • 4 years ago
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    NCERT

  5. anonymous
    • 4 years ago
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    For reference, you should not use the same variable for the upper limit as you do in the expression. Instead, say \[ \sum_{k=1}^n\space k^2 \]

  6. anonymous
    • 4 years ago
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    Do you mean prove it or derive it? It can be proved with induction.

  7. anonymous
    • 4 years ago
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    sorry i meant derive it?

  8. anonymous
    • 4 years ago
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    http://www.trans4mind.com/personal_development/mathematics/series/sumNaturalSquares.htm In the first proof I don't get this part "...Because \(Δ_3\) is a constant, the sum is a cubic of the form..."

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spraguer (Moderator)
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