The value of `i^(1+3+5++(2n+1))` is________.

Options

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

Explanation

To find the value of \( i^{(1 + 3 + 5 + \ldots + (2n + 1))} \), we will follow these steps:

### Step 1: Identify the series
The expression \( 1 + 3 + 5 + \ldots + (2n + 1) \) is the sum of the first \( n + 1 \) odd numbers.

### Step 2: Use the formula for the sum of odd numbers
The sum of the first \( n \) odd numbers is given by the formula:
\[ \text{Sum} = (n + 1)^2 \]
Thus, we can express our series as:
\[ 1 + 3 + 5 + \ldots + (2n + 1) = (n + 1)^2 \]

### Step 3: Substitute back into the expression
Now we substitute this result back into the original expression:
\[ i^{(1 + 3 + 5 + \ldots + (2n + 1))} = i^{(n + 1)^2} \]

### Step 4: Simplify \( i^{(n + 1)^2} \)
The powers of \( i \) cycle every 4:
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)

To find \( i^{(n + 1)^2} \), we need to determine \( (n + 1)^2 \mod 4 \).

### Step 5: Analyze \( (n + 1)^2 \mod 4 \)
- If \( n + 1 \) is even, then \( (n + 1)^2 \equiv 0 \mod 4 \) and \( i^{(n + 1)^2} = 1 \).
- If \( n + 1 \) is odd, then \( (n + 1)^2 \equiv 1 \mod 4 \) and \( i^{(n + 1)^2} = i \).

### Conclusion
Thus, we can summarize:
- If \( n + 1 \) is even (i.e., \( n \) is odd), then \( i^{(n + 1)^2} = 1 \).
- If \( n + 1 \) is odd (i.e., \( n \) is even), then \( i^{(n + 1)^2} = i \).

The final answer is:
\[ \text{The value of } i^{(1 + 3 + 5 + \ldots + (2n + 1))} \text{ is } \begin{cases} 1 & \text{if } n \text{ is odd} \\ i & \text{if } n \text{ is even} \end{cases} \]