smanujahrc wrote:

For a certain sequence \(a_1=9^1\) and \(a_n=(10^n - 1)^n\), where \(n\) is a positive integer. If S is the sum of \(n\) terms of the sequence, what is the units digit of S?

please can someone elaborate on the answer and explanation i cannot understand why cannot n be greater than 3.

i tried forming series first as

9,99^2, 999^3, 9999^4...... (10^n - 1)^n

how come series is in form ..9,..1,..9,..1 ??? what is this clearly ??

please help

That the terms of the sequence are given by the formula \(a_n=(10^n - 1)^n\).

\(a_1=9^1=9\)

\(a_2=99^2=...1\)

\(a_3=999^3=...9\)

\(a_4=9999^4=...1\)

...

As you can see odd terms have the units digit of 9 and the even terms have the units digit of 1.

The question asks about the units digit of the sum of \(n\) terms of the sequence. Now, if n is even then this sum ends with 0. If it's odd, then this sum ends with 9.

For example:

If n = 2, then the units digit of the sum of 2 terms is 9^1 + 99^2 = ...0.

If n = 3, then the units digit of the sum of 3 terms is 9^1 + 99^2 + 999^3 = ...9.

Statement (1) by itself is sufficient. n = 2 --> the sum ends with 0.

Statement (2) by itself is insufficient. (2 is prime and even, 3 is prime and odd).

Answer: A.

Hope it's clear.

Brilliant explanation. Thanx a lot. Its is superb.