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\[\Huge{1^2+2^2+3^2...........n^2=???}\]
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@waterineyes @UnkleRhaukus Plz help
\[\Huge{Ans.{n(n+1)(2n+1)\over6}}\]
Square pyramidal number
sorry \[\large (k+1)^3-k^3=3k^2+3k+1\] take sigma k=1 to n from both sides
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(k+1)^3=k^3+3k^2+3k+1 and we will havve (k+1)^3-k^3=3k^2+3k+1 a telescopic sum...(n+1)^3-0=3*\[\sum_{0}^{n}k^2\]+3*\[\sum_{0}^{n}k\]+\[\sum_{0}^{n}1\]
and as we know...\[\sum_{0}^{n}k=n(n+1)/2 and \sum_{0}^{n}1=n+1 \] you can finish it by yourself :)
no i want full sol @Neemo
\[okk (n+1)^3=3\sum_{0}^{n}k^2+3n(n+1)/2+n-1\] then \[3\sum_{0}^{n}k^2=(n+1)((n+1)^2-3n/2 -1)\] \[6\sum_{0}^{n}k^2=(n+1)(2n^2+4n+2-3n-2)\] \[6\sum_{0}^{n}k^2=(n+1)n(2n+1)\]
in the third line i multiplied both side by 2 ! It's clear now ?!
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