The value of `i^(2n)+i^(2n+1)+i^(2n+2)+i^(2n+3),` where `i=sqrt(-1),` is

Options

A. 0
B. 1
C. `i`
D. `-i`

Explanation

To solve the expression \( i^{2n} + i^{2n+1} + i^{2n+2} + i^{2n+3} \), where \( i = \sqrt{-1} \), we will first analyze the powers of \( i \).

### Step 1: Understand the powers of \( i \)
The powers of \( i \) cycle every 4 terms:
- \( i^0 = 1 \)
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \) (and the cycle repeats)

### Step 2: Determine \( i^{2n} \), \( i^{2n+1} \), \( i^{2n+2} \), and \( i^{2n+3} \)
Since the powers of \( i \) repeat every 4, we can express \( 2n \), \( 2n+1 \), \( 2n+2 \), and \( 2n+3 \) in terms of their residues modulo 4.

- \( 2n \mod 4 \) can be either \( 0 \) or \( 2 \) depending on whether \( n \) is even or odd.
- \( 2n+1 \mod 4 \) will be \( 1 \) if \( n \) is even and \( 3 \) if \( n \) is odd.
- \( 2n+2 \mod 4 \) will be \( 2 \) if \( n \) is even and \( 0 \) if \( n \) is odd.
- \( 2n+3 \mod 4 \) will be \( 3 \) if \( n \) is even and \( 1 \) if \( n \) is odd.

### Step 3: Evaluate the expression based on the parity of \( n \)

#### Case 1: \( n \) is even
- \( 2n \mod 4 = 0 \) → \( i^{2n} = 1 \)
- \( 2n+1 \mod 4 = 1 \) → \( i^{2n+1} = i \)
- \( 2n+2 \mod 4 = 2 \) → \( i^{2n+2} = -1 \)
- \( 2n+3 \mod 4 = 3 \) → \( i^{2n+3} = -i \)

So, the expression becomes:
\[ i^{2n} + i^{2n+1} + i^{2n+2} + i^{2n+3} = 1 + i - 1 - i = 0 \]

#### Case 2: \( n \) is odd
- \( 2n \mod 4 = 2 \) → \( i^{2n} = -1 \)
- \( 2n+1 \mod 4 = 3 \) → \( i^{2n+1} = -i \)
- \( 2n+2 \mod 4 = 0 \) → \( i^{2n+2} = 1 \)
- \( 2n+3 \mod 4 = 1 \) → \( i^{2n+3} = i \)

So, the expression becomes:
\[ i^{2n} + i^{2n+1} + i^{2n+2} + i^{2n+3} = -1 - i + 1 + i = 0 \]

### Conclusion
In both cases, whether \( n \) is even or odd, we find that:
\[ i^{2n} + i^{2n+1} + i^{2n+2} + i^{2n+3} = 0 \]

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