Official Solution:For a certain sequence \(a_1=9\) and \(a_n=(10^n - 1)^n\), where \(n\) is a positive integer. If S is the sum of the first 101 terms of the sequence, what is the units digit of S? A. 0
B. 1
C. 3
D. 7
E. 9
The terms of the sequence are given by the formula \(a_n = (10^n - 1)^n\).
\(a_1 = 9\)
\(a_2 = 99^2 = \dots 1\)
\(a_3 = 999^3 = \dots 9\)
\(a_4 = 9999^4 = \dots 1\)
...
As we can see, odd terms have the units digit of 9 and the even terms have the units digit of 1.
Now, if we add the first \(n\) terms where \(n\) is even, the sum ends with 0. If we add the first \(n\) terms where \(n\) is odd, this sum ends with 9.
For example:
If \(n = 2\), then the units digit of the sum of the first two terms is \(9 + 99^2 = 9+ ...1=\dots 0\).
If \(n = 3\), then the units digit of the sum of the first three terms is \(9 + 99^2 + 999^3 =9+...1+...9= \dots 9\).
The question asks about the units digit of the sum of the first 101, which is odd, terms of the sequence. Therefore, the units digit of \(S\) will be 9.
Answer: E