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

Question

jiteshmeghwal9 · Answer

@waterineyes @UnkleRhaukus Plz help

jiteshmeghwal9 · Answer

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

anonymous · Answer

Square pyramidal number

anonymous · Answer

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

unklerhaukus · Answer

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

anonymous · Answer

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

anonymous · Answer

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 :)

jiteshmeghwal9 · Answer

no i want full sol @Neemo

anonymous · Answer

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

anonymous · Answer

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