Question

(4 ^ (n + 1) * 2 ^ n - 8 ^ n)/(2 ^ (3n + 1))

Answer 1

Step-by-step explanation:

$\frac{{4}^{n + 1} \times 2 ^ n - 8 ^ n}{ {2}^{3n + 1} } \\ = \frac{{2}^{2n + 2} \times 2 ^ n - {2}^{3n} }{ {2}^{3n + 1} } \\ = \frac{ {2}^{2n + 2 + n} }{ {2}^{3n + 1} } - \frac{ {2}^{3n} }{ {2}^{3n + 1} } \\ = {2}^{3n + 2 - 3n - 1} - {2}^{3n - 3n - 1} \\ = 2 - {2}^{ - 1} \\ = 2 - \frac{1}{2} \\ = \frac{4 - 1}{2} \\ = \frac{3}{2}$

Answer 2

★GIVEN:-

• $= \frac{4 {}^{n + 1} \times 2 {}^{n} - 8 {}^{n} }{2 {}^{3n + 1} }$

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EVALAUTION★

$= \frac{4 {}^{n + 1} \times 2 {}^{n} - 8 {}^{n} }{2 {}^{3n + 1} }$

• can be this written as
• $= \frac{(2 \times 2 ){}^{n + 1} \times 2 {}^{ {}^{n} } -( 2 \times 2 \times 2) {}^{n} }{2 {}^{n + 1} } \\ \\ = \frac{2 {}^{n + 1} \times 2 {}^{n + 1} \times 2 {}^{n} - 2 {}^{n} \times 2 {}^{n} \times 2 {}^{n} }{2 {}^{n + 1} }$
• we know that

• $\frac{1}{a {}^{m} } = a {}^{m}$
• $= > a {}^{m} \times a {}^{n} = a {}^{m + n}$
• $= 2 {}^{n + l} \times 2 {}^{ - (n + 1)} \times 2 {}^{2n + 1} - 2 {}^{3n} \\ \\ = 2 {}^{n + 1 - n - 1} \times 2 {}^{2n} \times 2 - 2 {}^{2n} \times 2 {}^{n} \\ \\ = 2 {}^{0} \times 2 {}^{2n}(2 - 2 {}^{n}) \\ \\ = we \: know \\ \\ a {}^{0 } = 1 \\ \\ = 2 {}^{2n} (2 - 2 {}^{n} )$

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I hope it helps you