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Nov 8, 2020 at 21:52 comment added Gerry Myerson @Yaakov I'm happy either way. It seems OP had $n-1$ in mind.
Nov 8, 2020 at 16:01 comment added Yaakov Baruch @Gerry Myerson. Or $(n-1)b^2$ I thought was agreed on...
Nov 6, 2020 at 22:36 comment added Gerry Myerson But you want that last sum to come to $nb^2$, not $b^2$, no?
Nov 6, 2020 at 19:11 history answered Neil Strickland CC BY-SA 4.0