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 \(9\times 2^{2n}\)
waec 1999mathematics

The correct answer is C. \(9\times 2^{n-1}\)

\(K = 81 \times 2^{2n - 2}\)

\(\sqrt{K} = \sqrt{81 \times 2^{2n - 2}}\)

= \(9 \times 2^{n - 1}\)

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