Question

$\sum_{n=1}^{2025}i^{n}(1+i), i^{2}=-1$ is equal to ________.

• A. $i+1$
• B. $i-1$
• C. $-i-1$
• D. $-i+1$
• E. $-i$

Hint:

$i^1+i^2+i^3+i^4 = 0$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
We need to evaluate a summation involving powers of the imaginary unit \(i\). The powers of \(i\) are cyclical with a period of 4. We can use this property to simplify the summation. The term \((1+i)\) is a constant with respect to the summation index \(n\), so it can be factored out.
Step 2: Key Formula or Approach
The expression can be written as:
\[ (1+i) \sum_{n=1}^{2025} i^n \] The key is to evaluate the sum \(S = \sum_{n=1}^{2025} i^n = i^1 + i^2 + i^3 + \dots + i^{2025}\). The cycle of powers of \(i\) is: \(i^1=i\), \(i^2=-1\), \(i^3=-i\), \(i^4=1\).
The sum of any four consecutive powers is \(i - 1 - i + 1 = 0\).
Step 3: Detailed Explanation
1. Evaluate the summation \(\sum_{n=1}^{2025} i^n\).
We have a sum of 2025 terms. We can group these terms in sets of four. To find out how many full cycles of 4 we have and what the remainder is, we divide 2025 by 4.
\[ 2025 \div 4 = 506 \text{ with a remainder of } 1 \] So, we can write the sum as:
\[ S = (i^1 + i^2 + i^3 + i^4) + (i^5 + \dots + i^8) + \dots + (i^{2021} + \dots + i^{2024}) + i^{2025} \] There are 506 full groups of four terms. The sum of each group is 0.
\[ S = (0) + (0) + \dots + (0) + i^{2025} \] \[ S = 506 \times 0 + i^{2025} = i^{2025} \] Now we need to find the value of \(i^{2025}\). Since the remainder when 2025 is divided by 4 is 1,
\[ i^{2025} = i^{4 \times 506 + 1} = (i^4)^{506} \cdot i^1 = (1)^{506} \cdot i = i \] So, the sum is \(S = i\).
2. Multiply the result by \((1+i)\).
The original expression is \(S \times (1+i)\).
\[ i \times (1+i) = i(1) + i(i) = i + i^2 \] Since \(i^2 = -1\), we have:
\[ i - 1 \] Step 4: Final Answer
The value of the expression is \(i-1\).