The units' digit of the sum `1+9+9^(2)+......9^(1006)` is

Options

A. 2
B. 1
C. 9
D. 0

Answer

To find the units digit of the sum \(1 + 9 + 9^2 + \ldots + 9^{1006}\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Pattern of Units Digits:**
- First, we need to find the units digits of the powers of 9.
- Calculate the first few powers of 9:
- \(9^1 = 9\) (units digit is 9)
- \(9^2 = 81\) (units digit is 1)
- \(9^3 = 729\) (units digit is 9)
- \(9^4 = 6561\) (units digit is 1)

From this, we can see that the units digits of the powers of 9 alternate between 9 and 1:
- Odd powers of 9 have a units digit of 9.
- Even powers of 9 have a units digit of 1.

2. **List the Terms:**
- The sum can be rewritten as:
\[ 1 + 9 + 9^2 + 9^3 + \ldots + 9^{1006} \]
- This includes 1007 terms (from \(9^0\) to \(9^{1006}\)).

3. **Count Odd and Even Terms:**
- The first term \(1\) (which is \(9^0\)) has a units digit of 1.
- The sequence continues with:
- 1 (from \(9^0\))
- 9 (from \(9^1\))
- 1 (from \(9^2\))
- 9 (from \(9^3\))
- ...
- The pattern continues, and we can see that:
- Odd indexed terms (1st, 3rd, 5th, ...) contribute 1.
- Even indexed terms (2nd, 4th, 6th, ...) contribute 9.

4. **Calculate the Total Contribution:**
- There are 1007 terms in total:
- Odd indexed terms: \(9^0, 9^2, 9^4, \ldots, 9^{1006}\) (504 terms contributing 1)
- Even indexed terms: \(9^1, 9^3, 9^5, \ldots, 9^{1005}\) (503 terms contributing 9)
- Calculate the contributions:
- Contribution from odd indexed terms: \(504 \times 1 = 504\)
- Contribution from even indexed terms: \(503 \times 9 = 4527\)

5. **Sum the Contributions:**
- Total contribution = \(504 + 4527 = 5031\)

6. **Find the Units Digit:**
- The units digit of \(5031\) is **1**.

### Final Answer:
The units digit of the sum \(1 + 9 + 9^2 + \ldots + 9^{1006}\) is **1**.