smc

To solve this problem, we will be using difference of cube, sum of squares and sum of arithmetic sequence formulas. $$2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3$$ $=(2-1)(2^2+1\cdot 2+1^2)+(4-3)(4^2+4\cdot 3+3^2)+(6-5)(6^2+6\cdot 5+5^2)+...+(18-17)(18^2+18\cdot 17+17^2)$ $=(2^2+1\cdot 2+1^2)+(4^2+4\cdot 3+3^2)+(6^2+6\cdot 5+5^2)+...+(18^2+18\cdot 17+17^2)$ $=1^2+2^2+3^2+4^2+5^2+6^2...+17^2+18^2+1\cdot 2+3\cdot 4+5\cdot 6+...+17\cdot 18$ $=\frac{18(18+1)(36+1)}{6}+1\cdot 2+3\cdot 4+5\cdot 6+...+17\cdot 18$ we could rewrite the second part as ${\sum }_{n=1}^9(2n-1)(2n)$ $(2n-1)(2n)=4n^2-2n$ ${\sum }_{n=1}^{9}4n^2=4(\frac{9(9+1)(18+1)}{6})$ ${\sum }_{n=1}^9-2n=-2(\frac{9(9+1)}{2})$ Hence, $1\cdot 2+3\cdot 4+5\cdot 6+...+17\cdot 18=4(\frac{9(9+1)(18+1)}{6})-2(\frac{9(9+1)}{2})$ Adding everything up: $2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3$ $=\frac{18(18+1)(36+1)}{6}+4(\frac{9(9+1)(18+1)}{6})-2(\frac{9(9+1)}{2})$ $=3(19)(37)+6(10)(19)-9(10)$ $=2109+1140-90$ $=\boxed{3159}$