KCET · Maths · Complex Number
\(\sum\limits_{n=1}^{4} (-1)^{2n} \cdot i^{2n} = \)
- A \(2\)
- B \(-i\)
- C \(0\)
- D \(i\)
Answer & Solution
Correct Answer
(C) \(0\)
Step-by-step Solution
Detailed explanation
The given expression is \(\sum\limits_{n=1}^{4} (-1)^{2n} \cdot i^{2n}\)
Since \((-1)^{2n} = ((-1)^2)^n = 1^n = 1\) for all integers \(n\), and \(i^{2n} = (i^2)^n = (-1)^n\), the expression simplifies to:
\(\sum\limits_{n=1}^{4} 1 \cdot (-1)^n = \sum\limits_{n=1}^{4} (-1)^n\)
Expanding the sum for \(n = 1, 2, 3, 4\):
\((-1)^1 + (-1)^2 + (-1)^3 + (-1)^4\)
\(-1 + 1 - 1 + 1 = 0\)
Answer: \(0\)
Since \((-1)^{2n} = ((-1)^2)^n = 1^n = 1\) for all integers \(n\), and \(i^{2n} = (i^2)^n = (-1)^n\), the expression simplifies to:
\(\sum\limits_{n=1}^{4} 1 \cdot (-1)^n = \sum\limits_{n=1}^{4} (-1)^n\)
Expanding the sum for \(n = 1, 2, 3, 4\):
\((-1)^1 + (-1)^2 + (-1)^3 + (-1)^4\)
\(-1 + 1 - 1 + 1 = 0\)
Answer: \(0\)
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