- #1
transgalactic
- 1,395
- 0
[tex]
1^2+2^2+..+k^2=\frac{k(k+1)(k+2)}{6}
[/tex]
how to get this general result
??
i know
1+2+..+n=\frac{n(n+1)}{2}
so by that rule the sum should be
[tex]
\frac{n(n^2+1)}{2}
[/tex]
1^2+2^2+..+k^2=\frac{k(k+1)(k+2)}{6}
[/tex]
how to get this general result
??
i know
1+2+..+n=\frac{n(n+1)}{2}
so by that rule the sum should be
[tex]
\frac{n(n^2+1)}{2}
[/tex]