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# help

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106
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Find the units digits of \(1 + 9 + 9^2 + 9^3 + \dots + 9^{100}\)

Jul 6, 2020

#1
0

sumfor(n, 0, 100, 9^n) =2.988157375 E+95 - This number ends in 501

Jul 6, 2020
#2
+31340
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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.

Jul 6, 2020