Simplifiy
`i^(2)+i^(4)+i^(6)+.....+(2n+1)` terms

Answer

To simplify the expression \( i^2 + i^4 + i^6 + \ldots + i^{2n} \) where there are \( n + 1 \) terms, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the pattern of powers of \( i \)**:
The powers of \( i \) cycle every four terms:
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)
- \( i^5 = i \)
- \( i^6 = -1 \)
- \( i^7 = -i \)
- \( i^8 = 1 \)
- And so on...

2. **Determine the powers in the series**:
The series consists of even powers of \( i \):
- \( i^2, i^4, i^6, \ldots, i^{2n} \)

3. **Calculate the values of the relevant powers**:
- For even powers:
- \( i^2 = -1 \)
- \( i^4 = 1 \)
- \( i^6 = -1 \)
- \( i^8 = 1 \)
- Continuing this pattern, we see that:
- \( i^{2k} = -1 \) for odd \( k \)
- \( i^{2k} = 1 \) for even \( k \)

4. **Count the number of terms**:
The series has \( n + 1 \) terms, which means:
- If \( n \) is even, there are \( \frac{n + 1}{2} \) terms equal to \( -1 \) and \( \frac{n + 1}{2} \) terms equal to \( 1 \).
- If \( n \) is odd, there are \( \frac{n + 2}{2} \) terms equal to \( -1 \) and \( \frac{n - 1}{2} \) terms equal to \( 1 \).

5. **Sum the series based on the counts**:
- If \( n \) is even:
\[ \text{Sum} = \left(\frac{n + 1}{2}\right)(1) + \left(\frac{n + 1}{2}\right)(-1) = 0 \]
- If \( n \) is odd:
\[ \text{Sum} = \left(\frac{n + 2}{2}\right)(-1) + \left(\frac{n - 1}{2}\right)(1) = -\frac{n + 2}{2} + \frac{n - 1}{2} = -\frac{3}{2} \]

6. **Final Result**:
- For even \( n \), the sum is \( 0 \).
- For odd \( n \), the sum is \( -\frac{3}{2} \).

### Conclusion:
The simplified result of the series \( i^2 + i^4 + i^6 + \ldots + i^{2n} \) depends on whether \( n \) is even or odd.