If `a + ib = sum_(k = 1)^(101)i^(k)`, then (a, b) equals

Options

A. (0, 1)
B. (0, 0)
C. (0, -1)
D. (1, 1)

Answer

To solve the problem \( a + ib = \sum_{k=1}^{101} i^k \), we will follow these steps:

### Step 1: Understand the powers of \( i \)
The powers of \( i \) (the imaginary unit) cycle every four terms:
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)
- \( i^5 = i \) (and so on...)

### Step 2: Determine the number of complete cycles in the summation
Since the powers of \( i \) repeat every four terms, we can find how many complete cycles fit into the sum from \( k = 1 \) to \( k = 101 \):
- The number of complete cycles of 4 in 101 is \( \left\lfloor \frac{101}{4} \right\rfloor = 25 \) complete cycles, with a remainder of \( 1 \).

### Step 3: Calculate the contribution from complete cycles
Each complete cycle contributes:
\[ i + (-1) + (-i) + 1 = 0 \]
Since there are 25 complete cycles, their total contribution is:
\[ 25 \times 0 = 0 \]

### Step 4: Calculate the contribution from the remaining terms
The remainder when dividing 101 by 4 is 1, which means we only have \( i^1 \) left to add:
\[ i^1 = i \]

### Step 5: Combine contributions
Adding the contributions from the complete cycles and the remaining term:
\[ \sum_{k=1}^{101} i^k = 0 + i = i \]

### Step 6: Identify real and imaginary parts
Now we have:
\[ a + ib = i \]
This means:
- The real part \( a = 0 \)
- The imaginary part \( b = 1 \)

### Final Answer
Thus, the values of \( (a, b) \) are:
\[ \boxed{(0, 1)} \]