The last digit of the expression
`4 + 9^2 + 4^3 + 9^4 + 4^5 + 9^6 + …..+ 4^(99) + 9^(100)` is:

Options

A. 0
B. 3
C. 5
D. none of these

Answer

To find the last digit of the expression \( 4 + 9^2 + 4^3 + 9^4 + 4^5 + 9^6 + \ldots + 4^{99} + 9^{100} \), we can break it down into two separate series: one for the powers of 4 and one for the powers of 9.

### Step 1: Identify the terms
The expression consists of alternating terms of \( 4 \) and \( 9 \):
- The terms with \( 4 \) are: \( 4^1, 4^3, 4^5, \ldots, 4^{99} \)
- The terms with \( 9 \) are: \( 9^2, 9^4, 9^6, \ldots, 9^{100} \)

### Step 2: Count the number of terms
- The powers of \( 4 \) (odd powers) range from \( 1 \) to \( 99 \). This gives us \( 50 \) terms.
- The powers of \( 9 \) (even powers) range from \( 2 \) to \( 100 \). This also gives us \( 50 \) terms.

### Step 3: Determine the last digit of the powers of 4
The last digits of the powers of \( 4 \) cycle every two terms:
- \( 4^1 \) has a last digit of \( 4 \)
- \( 4^2 \) has a last digit of \( 6 \)
- \( 4^3 \) has a last digit of \( 4 \)
- \( 4^4 \) has a last digit of \( 6 \)
- Thus, for odd powers, the last digit is always \( 4 \).

Since there are \( 50 \) terms of \( 4 \) (which are all odd powers), the contribution to the last digit from the \( 4 \) terms is:
\[ 50 \times 4 = 200 \]
The last digit of \( 200 \) is \( 0 \).

### Step 4: Determine the last digit of the powers of 9
The last digits of the powers of \( 9 \) also cycle every two terms:
- \( 9^1 \) has a last digit of \( 9 \)
- \( 9^2 \) has a last digit of \( 1 \)
- \( 9^3 \) has a last digit of \( 9 \)
- \( 9^4 \) has a last digit of \( 1 \)
- Thus, for even powers, the last digit is always \( 1 \).

Since there are \( 50 \) terms of \( 9 \) (which are all even powers), the contribution to the last digit from the \( 9 \) terms is:
\[ 50 \times 1 = 50 \]
The last digit of \( 50 \) is \( 0 \).

### Step 5: Combine the last digits
Now, we add the last digits from both contributions:
\[ \text{Last digit of } (4 \text{ terms}) + \text{Last digit of } (9 \text{ terms}) = 0 + 0 = 0 \]

### Final Answer
Thus, the last digit of the entire expression \( 4 + 9^2 + 4^3 + 9^4 + 4^5 + 9^6 + \ldots + 4^{99} + 9^{100} \) is \( \boxed{0} \).