Question

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Answer 1

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Answer 2

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$\frac{ {4}^{n + 1}. {2}^{n} - {8}^{n} }{ {2}^{3m} } = \frac{3}{8} \\ or. \: \frac{ {2}^{2(n + 1)}. {2}^{n} - {2}^{3n} }{ {2}^{3m} } = \frac{3}{8} \\ or. \: \frac{ {2}^{2(n + 1) + n} - {2}^{3n} }{ {2}^{3m} } = \frac{3}{ {2}^{3} } \\ or. \: \frac{ {2}^{3n}(4 - 1) }{ {2}^{3m} } = \frac{3}{ {2}^{3} } \\ or \frac{ {2}^{3n} \times 3}{ {2}^{3m} } = \frac{ {2}^{0} \times 3 }{ {2}^{3} } \\ \\ so.. \: \: comparing \: we \: get \: n = 0 \: \: and \: \: m = 1 \\ thus.. \\ n + 1 = m.....proved....$
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