Given that \(81\times 2^{2n-2} = K, find \sqrt{K}\)
MATHEMATICS
WAEC 1999
Given that \(81\times 2^{2n-2} = K, find \sqrt{K}\)
- A. \(4.5\times 2^{n}\)
- B. \(4.5\times 2^{2n}\)
- C. \(9\times 2^{n-1}\)
- D. \(9\times 2^{2n}\)
Correct Answer: C. \(9\times 2^{n-1}\)
Explanation
\(K = 81 \times 2^{2n - 2}\)
\(\sqrt{K} = \sqrt{81 \times 2^{2n - 2}}\)
= \(9 \times 2^{n - 1}\)
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