I had to look this one up, but I knew there was an equation:
the sum of squares of first 'n' natural numbers:
1^2+2^2+3^2+.....+n^2=n*(n+1)*(2n+1)/6
the sum of cubes of first 'n' natural numbers:
1^3+2^3+3^3+4^3+......+n^3=(n^2 * (n+1)^2)/4
I'm not sure where these equations came from, but they can be proved by induction.
singingbanana 1 year ago
@singingbanana There it is. I have been trying to figure that out for the past week. I can't thank you enough because it has been driving me crazy. Now I may give a try in finding how this equation was found. Thank you
PEZenfuego 1 year ago
I had to look this one up, but I knew there was an equation:
the sum of squares of first 'n' natural numbers:
1^2+2^2+3^2+.....+n^2=n*(n+1)*(2n+1)/6
the sum of cubes of first 'n' natural numbers:
1^3+2^3+3^3+4^3+......+n^3=(n^2 * (n+1)^2)/4
I'm not sure where these equations came from, but they can be proved by induction.
singingbanana 1 year ago
@singingbanana There it is. I have been trying to figure that out for the past week. I can't thank you enough because it has been driving me crazy. Now I may give a try in finding how this equation was found. Thank you
PEZenfuego 1 year ago