Question

If \( i = \sqrt{-1} \), then \[ \sum_{n=2}^{30} i^n + \sum_{n=30}^{65} i^{n+3} = \]

• A. \(0\)
• B. \(-1\)
• C. \(i\)
• D. \(-i\)

Hint:

Remember powers of \(i\) cycle every 4: \(i, -1, -i, 1\).

Answer 1

The Correct Option is B

Step 1: Recall powers of \(i\)
\[ i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad \text{and then repeats every 4 powers} \] Step 2: Compute \( S_1 = \sum_{n=2}^{30} i^n \)
Since powers of \( i \) repeat every 4, sum the terms modulo 4. Number of terms: \( 30 - 2 + 1 = 29 \). Sum of one full cycle (4 terms): \( i^1 + i^2 + i^3 + i^4 = i -1 - i + 1 = 0 \). Adjust starting power and count terms accordingly to calculate. Step 3: Compute \( S_2 = \sum_{n=30}^{65} i^{n+3} = \sum_{m=33}^{68} i^m \)
Similarly, sum the powers \( i^{33} \) to \( i^{68} \). Step 4: Combine sums and simplify
By calculating modulo 4 cycles and simplifying, the total sum equals \(-1\).