Question

\[\Huge{1^2+2^2+3^2...........n^2=???}\]

Answer 1

@waterineyes @UnkleRhaukus Plz help

Answer 2

\[\Huge{Ans.{n(n+1)(2n+1)\over6}}\]

Answer 3

Square pyramidal number

Answer 4

sorry
\[\large (k+1)^3-k^3=3k^2+3k+1\]
take sigma k=1 to n from both sides

Answer 5

http://books.google.com.au/books?id=cyyhZr-SffcC&lpg=PA67&dq=proofs+without+words+sum+of+cubes&pg=PA85&redir_esc=y#v=onepage&q&f=false

Answer 6

(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*\[\sum_{0}^{n}k\]+\[\sum_{0}^{n}1\]

Answer 7

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 :)

Answer 8

no i want full sol @Neemo

Answer 9

\[okk (n+1)^3=3\sum_{0}^{n}k^2+3n(n+1)/2+n-1\]
then \[3\sum_{0}^{n}k^2=(n+1)((n+1)^2-3n/2 -1)\]
\[6\sum_{0}^{n}k^2=(n+1)(2n^2+4n+2-3n-2)\]
\[6\sum_{0}^{n}k^2=(n+1)n(2n+1)\]

Answer 10

in the third line i multiplied both side by 2 ! It's clear now ?!