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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. across
    • 3 years ago
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    This could follow from Gauss' assertion that\[\sum_{i=1}^{n}i=\frac{n(n+1)}{2}\]

  2. helder_edwin
    • 3 years ago
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    i don't remember the details. but i'm pretty sure u can find this in Spivaks' Calculus

  3. across
    • 3 years ago
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    It's easy to prove this by induction with your eyes closed, but I'm wondering how it's derived.

  4. waterineyes
    • 3 years ago
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    Can I provide the link or not here ??

  5. waterineyes
    • 3 years ago
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    @across can I provide here the link or I have to derive here full??

  6. experimentX
    • 3 years ago
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    i think this was done in physics section ...

  7. waterineyes
    • 3 years ago
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    This is nothing but sum of squares of first n natural numbers..

  8. helder_edwin
    • 3 years ago
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    yes i know the proof is easy. i suggested spivak's book because i remember seeing the derivation there

  9. experimentX
    • 3 years ago
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    i would prefer geometrical visualization instead ... i remember seeing one is mit ocw single variable calculus. the volume of pyramid ...

  10. across
    • 3 years ago
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    Links are more appreciated than derivations here. ^^

  11. waterineyes
    • 3 years ago
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    http://www.ilovemaths.com/3sequence.asp May be it will help you.

  12. waterineyes
    • 3 years ago
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    http://www.9math.com/book/sum-squares-first-n-natural-numbers Or you can check this also..

  13. experimentX
    • 3 years ago
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    difference of power is very power technique ... it can be used to find the sums of n^3 and n^4 also ....

  14. asnaseer
    • 3 years ago
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    There is a very beautiful visual proof of this that @Ishaan94 came across when he asked this question: http://openstudy.com/updates/5002f7dae4b0848ddd66eea4

  15. experimentX
    • 3 years ago
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    very interesting proof indeed !! \[ 3 \left ( \sum_{n=1}^\infty n^2 \right ) = (1 + 2 +3 +... + n)(2n+1)\]

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