`i^(13)+i^(18)+i^(31)+n=0`
In the equation above, what is the value of n in simplest form?

Options

A. `-i`
B. `-1`
C. `1`
D. `i`

Answer

To solve the equation \( i^{13} + i^{18} + i^{31} + n = 0 \), we need to evaluate the powers of \( i \) involved in the equation.

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

### Step 2: Calculate \( i^{13} \)
To find \( i^{13} \), we can use the property of the cycle:
\[ 13 \mod 4 = 1 \]
Thus, \( i^{13} = i^1 = i \).

### Step 3: Calculate \( i^{18} \)
Next, for \( i^{18} \):
\[ 18 \mod 4 = 2 \]
Thus, \( i^{18} = i^2 = -1 \).

### Step 4: Calculate \( i^{31} \)
Now for \( i^{31} \):
\[ 31 \mod 4 = 3 \]
Thus, \( i^{31} = i^3 = -i \).

### Step 5: Substitute back into the equation
Now we can substitute these values back into the original equation:
\[ i + (-1) + (-i) + n = 0 \]
This simplifies to:
\[ i - i - 1 + n = 0 \]
Which further simplifies to:
\[ -1 + n = 0 \]

### Step 6: Solve for \( n \)
To isolate \( n \), we add 1 to both sides:
\[ n = 1 \]

### Conclusion
The value of \( n \) in simplest form is:
\[ \boxed{1} \]