Question:
The integer just greater than \((3 + \sqrt{5})^{2n}\) is divisible by
The integer just greater than \((3 + \sqrt{5})^{2n}\) is divisible by
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Use conjugates to evaluate expressions with irrational powers.
Updated On: Mar 23, 2026
- \(2^{\,n-1}\)
- \(2^{\,n+1}\)
- \(2^{\,n+2}\)
- Not divisible by 2
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The Correct Option is B
Solution and Explanation
Step 1: Expand using the binomial theorem:
\[ (3 + \sqrt{5})^{2n} + (3 - \sqrt{5})^{2n} \] is an integer.
Step 2: Since \(0 < (3 - \sqrt{5})^{2n} < 1\),
\[ \lceil (3 + \sqrt{5})^{2n} \rceil = (3 + \sqrt{5})^{2n} + (3 - \sqrt{5})^{2n} \]
Step 3: The sum is divisible by \(2^n + 1\).
\[ (3 + \sqrt{5})^{2n} + (3 - \sqrt{5})^{2n} \] is an integer.
Step 2: Since \(0 < (3 - \sqrt{5})^{2n} < 1\),
\[ \lceil (3 + \sqrt{5})^{2n} \rceil = (3 + \sqrt{5})^{2n} + (3 - \sqrt{5})^{2n} \]
Step 3: The sum is divisible by \(2^n + 1\).
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