The value of `1/(i^n) + 1/(i^(n + 3)) + 1/(i^(n + 2)) + 1/(i^(n + 1))` is :

Options

A. `-1`
B. `1`
C. `0`
D. can't be determined

Answer

To solve the expression \( \frac{1}{i^n} + \frac{1}{i^{n+1}} + \frac{1}{i^{n+2}} + \frac{1}{i^{n+3}} \), we can simplify each term step by step.

### Step 1: Rewrite the terms using properties of exponents
We know that \( i = \sqrt{-1} \) and that \( i^2 = -1 \). The powers of \( i \) cycle every four terms:
- \( i^0 = 1 \)
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \) (and so on)

Thus, we can express each term in the sum:
\[ \frac{1}{i^n} + \frac{1}{i^{n+1}} + \frac{1}{i^{n+2}} + \frac{1}{i^{n+3}} = i^{-n} + i^{-(n+1)} + i^{-(n+2)} + i^{-(n+3)} \]

### Step 2: Factor out \( i^{-n} \)
We can factor out \( i^{-n} \) from the expression:
\[ = i^{-n} \left( 1 + i^{-1} + i^{-2} + i^{-3} \right) \]

### Step 3: Simplify the terms inside the parentheses
Now, we need to evaluate \( 1 + i^{-1} + i^{-2} + i^{-3} \). We know:
- \( i^{-1} = \frac{1}{i} = -i \) (since \( i \cdot -i = 1 \))
- \( i^{-2} = \frac{1}{i^2} = -1 \)
- \( i^{-3} = \frac{1}{i^3} = i \)

Substituting these values in:
\[ 1 + (-i) + (-1) + i = 1 - 1 + (-i + i) = 0 \]

### Step 4: Combine the results
Thus, we have:
\[ i^{-n} \cdot 0 = 0 \]

### Final Answer
The value of the expression \( \frac{1}{i^n} + \frac{1}{i^{n+1}} + \frac{1}{i^{n+2}} + \frac{1}{i^{n+3}} \) is \( \boxed{0} \).

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