Question

Evaluate the sum \[ \sum_{n=1}^{24} \left( i^n + i^{n+1} \right) \] is:

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

Hint:

When dealing with powers of \( i \), recognize the periodicity every four terms. This can simplify evaluating sums or expressions involving powers of \( i \).

Answer 1

The Correct Option is D

We are asked to evaluate the sum:

\[ \sum_{n=1}^{24} \left( i^n + i^{n+1} \right). \] Step 1: Use the periodicity of powers of \( i \). The powers of \( i \) repeat every four terms:

\[ i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad \text{and then the cycle repeats.} \] Thus, the powers of \( i \) repeat every four terms.

Step 2: Notice that \( i^n + i^{n+1} \) will follow a periodic pattern as well. Let's compute the first few terms to identify the pattern:

- For \( n = 1 \), \( i^1 + i^2 = i + (-1) = i - 1 \),
- For \( n = 2 \), \( i^2 + i^3 = -1 + (-i) = -1 - i \),
- For \( n = 3 \), \( i^3 + i^4 = -i + 1 = 1 - i \),
- For \( n = 4 \), \( i^4 + i^5 = 1 + i = 1 + i \).

Step 3: The sum \( i^n + i^{n+1} \) follows the pattern \( i - 1, -1 - i, 1 - i, 1 + i \), which repeats every four terms.

Step 4: Since the sum repeats every 4 terms and we are summing from \( n = 1 \) to \( n = 24 \), we have 6 full cycles of the 4-term pattern. The sum of one full cycle is:

\[ (i - 1) + (-1 - i) + (1 - i) + (1 + i) = 0. \] Thus, each cycle contributes 0 to the sum.

Step 5: Since there are 6 full cycles, the total sum is:

\[ 6 \times 0 = 0. \] Thus, the value of the sum is \( 0 \).

Therefore, the correct answer is option (D).