<p>The value of \(i^{1+3+5+\cdots+(2n+1)}\) is ________.</p>
Step-by-Step Solution
Key Concept: The exponent is the sum of the first (n+1) odd numbers, which equals (n+1)². Since i⁴ = 1, only the remainder when (n+1)² is divided by 4 matters.
<p><strong>Step 1:</strong> Find the sum of the exponent. The series 1+3+5+···+(2n+1) is the sum of the first (n+1) odd natural numbers.</p><p><strong>Step 2:</strong> Use the formula: sum of first m odd numbers = m². Here m = n+1, so the sum = (n+1)²</p><p><strong>Step 3:</strong> We need i^((n+1)²). Since i¹=i, i²=-1, i³=-i, i⁴=1, and the pattern repeats every 4 powers, we find (n+1)² mod 4.</p><p><strong>Step 4:</strong> If n is odd, then n+1 is even. Let n+1=2k. Then (n+1)²=4k²≡0 (mod 4), so i^((n+1)²) = i⁰ = 1</p><p><strong>Step 5:</strong> If n is even, then n+1 is odd. Let n+1=2k+1. Then (n+1)²=4k²+4k+1≡1 (mod 4), so i^((n+1)²) = i¹ = i</p><p>∴ <strong>Answer:</strong> 1 if n is odd; i if n is even</p>
Correct Answer: 1 if n is odd, i if n is even