Question:
$\sum_{n=1}^{2025}i^{n}(1+i), i^{2}=-1$ is equal to ________.
$\sum_{n=1}^{2025}i^{n}(1+i), i^{2}=-1$ is equal to ________.
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$i^1+i^2+i^3+i^4 = 0$.
Updated On: Jun 26, 2026
- $i+1$
- $i-1$
- $-i-1$
- $-i+1$
- $-i$
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Concept
Sum of four consecutive powers of $i$ is zero.
Step 2: Meaning
$\sum_{n=1}^{2025}i^n = (i^1 + i^2 + i^3 + i^4) + ... + i^{2025}$.
Step 3: Analysis
$2025 \pmod 4 = 1$, so the sum reduces to $i^1 = i$.
Step 4: Conclusion
The expression is $i(1+i) = i + i^2 = i - 1$. Final Answer: (B)
Sum of four consecutive powers of $i$ is zero.
Step 2: Meaning
$\sum_{n=1}^{2025}i^n = (i^1 + i^2 + i^3 + i^4) + ... + i^{2025}$.
Step 3: Analysis
$2025 \pmod 4 = 1$, so the sum reduces to $i^1 = i$.
Step 4: Conclusion
The expression is $i(1+i) = i + i^2 = i - 1$. Final Answer: (B)
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