Find the units digits of \(1 + 9 + 9^2 + 9^3 + \dots + 9^{100}\)
sumfor(n, 0, 100, 9^n) =2.988157375 E+95 - This number ends in 501
Every even power of 9 ends in 1, every odd power ends in 9. There are 51 even powers (including 90) and 50 odd powers.
Summing these we get 51*1 + 50*9 = 501, so the sum ends in 1.
