If ` i = sqrt(-1) ` and n is a positive integer, then ` i^(n) + i^(n + 1) + i^(n + 2) + i^(n + 3)`

Options

A. 1
B. i
C. `i^(n)`
D. 0

Answer

To solve the expression \( i^n + i^{n+1} + i^{n+2} + i^{n+3} \), where \( i = \sqrt{-1} \) and \( n \) is a positive integer, we can follow these steps:

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

### Step 2: Rewrite the expression
We can rewrite the expression \( i^n + i^{n+1} + i^{n+2} + i^{n+3} \) by factoring out \( i^n \):
\[ i^n + i^{n+1} + i^{n+2} + i^{n+3} = i^n (1 + i + i^2 + i^3) \]

### Step 3: Calculate \( 1 + i + i^2 + i^3 \)
Now, we need to calculate \( 1 + i + i^2 + i^3 \):
- We know \( i^2 = -1 \) and \( i^3 = -i \).
Thus,
\[ 1 + i + (-1) + (-i) = 1 - 1 + i - i = 0 \]

### Step 4: Substitute back into the expression
Substituting back, we have:
\[ i^n (1 + i + i^2 + i^3) = i^n \cdot 0 = 0 \]

### Conclusion
Therefore, the final result is:
\[ i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0 \]