Find the remainder when the sum 2^{2} + 2^{6} + 2^{10} + 2^{14} + \cdots + 2^{110} + 2^{114} + 2^{118} is divided by 10.
sumfor(n, 1, 30, 2^(4*n - 2))==354460 7988759775 6610768188 2741425220 mod 10 ==0 - the remainder.
+874 2^2 mod 10 = 4
2^6 mod 10 = 4
2^10 mod 10 = 4
any 2^(4k+2) mod 10 = 4, k≥0, can you see why?
(118-2)/4 + 1 = 30
30 * 4 mod 10 = 120 mod 10 = 0 mod 10.
answer is 0
