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across
 3 years ago
Let's go over the derivation of\[\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}\]
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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ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Mathematics group. Uhm.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Does this follow this intuition, somewhat? \[ \sum_{i = 1}^{n} 2i  1= n^2\]?

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.2\[(k+1)^3=k^3+3k^2+3k+1 \ \ or \ \ (k+1)^3k^3=3k^2+3k+1\] apply sum from \(1\) to \(n\) \[(n+1)^31=\sum_{k=1}^{n} [(k+1)^3k^3]=\sum_{k=1}^{n} [3k^2+3k+1]=3\sum_{k=1}^{n} k^2+3\frac{n(n+1)}{2}+n\]

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.2oh this is physics group...o.O.....
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