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minus_twelve_._note [2019/09/09 22:33] nikolaj |
minus_twelve_._note [2019/09/09 22:45] nikolaj |
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| So as an example, with $k=3,a=2,b=4$ you get the identity | So as an example, with $k=3,a=2,b=4$ you get the identity | ||
| - | $$\frac{1}{3}(4^3-2^3)=(2^2+3^2)+\frac{1}{2}(4^2-2^2)-\frac{1}{12}\,2\,(4^1-2^1)$$ | + | $$\frac{1}{3}(4^3-2^3)=(2^2+(4-1)^2)+\frac{1}{2}(4^2-2^2)-\frac{1}{12}\,2\,(4^1-2^1)$$ |
| and there are literally infinitely many identities involving $-\frac{1}{12}$ because of this formula. In case you're wondering, both sides of the equation above simplify to $\tfrac{56}{3}$. | and there are literally infinitely many identities involving $-\frac{1}{12}$ because of this formula. In case you're wondering, both sides of the equation above simplify to $\tfrac{56}{3}$. | ||
