- #1
pyfgcr
- 22
- 0
Hi everyone. I have learned that:
1+2+3+...=[itex]\frac{n(n+1)}{2}[/itex]
12+22+32=[itex]\frac{n(n+1)(2n+1)}{6}[/itex]
I want to know what the general formula of Ʃna, in which n and a are natural numbers, respect to n and a.
1+2+3+...=[itex]\frac{n(n+1)}{2}[/itex]
12+22+32=[itex]\frac{n(n+1)(2n+1)}{6}[/itex]
I want to know what the general formula of Ʃna, in which n and a are natural numbers, respect to n and a.