Hi all,
I got one question here.
Find the units digit in the sum $\displaystyle 1^{3}-2^{3}+3^{3}-4^{3}+...+39^{3}$
Thanks!
Difference of two cubes, maybe ? You could look at your expression like this :
$\displaystyle (1^3 - 2^3) + (3^3 - 4^3) + ... + (37^3 - 38^3) + 39^3$
Does that make sense ?
Or maybe by considering modular arithmetic modulo $\displaystyle 10$, which would conservate only the last digit ...