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across

  • 3 years ago

Let's go over the derivation of\[\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}\]

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  1. ParthKohli
    • 3 years ago
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    Mathematics group. Uhm.

  2. ParthKohli
    • 3 years ago
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    Does this follow this intuition, somewhat? \[ \sum_{i = 1}^{n} 2i - 1= n^2\]?

  3. mukushla
    • 3 years ago
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    \[(k+1)^3=k^3+3k^2+3k+1 \ \ or \ \ (k+1)^3-k^3=3k^2+3k+1\] apply sum from \(1\) to \(n\) \[(n+1)^3-1=\sum_{k=1}^{n} [(k+1)^3-k^3]=\sum_{k=1}^{n} [3k^2+3k+1]=3\sum_{k=1}^{n} k^2+3\frac{n(n+1)}{2}+n\]

  4. mukushla
    • 3 years ago
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    oh this is physics group...o.O.....

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