Question

please
$( { {2}^{ - 1} + {4}^{ - 1} + {6}^{ - 1 } + {8}^{ - 1} })^{n} = 1$
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Answer 1

Step-by-step explanation:

${({2}^{ - 1} + {4}^{ - 1} + {6}^{ - 1} + {8}^{ - 1})}^{n} = 1 \\ = > {(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8})}^{n} = 1 \\ = > {( \frac{12 + 6 + 4 + 3}{24}) }^{n} = 1 \\ = > {(\frac{25}{24})}^{n} = 1 \\ Now ,\: we \: know \: any \: number\: with \\ power \: zero = 1 \\ So ,\: we \: put \:( n )\: as \: 0. \\ \\ = > { (\frac{25}{24} )}^{0} = 1 \\ = > 1 = 1 \\ \\ Therefore \: \: \: LHS = RHS\\ \\ Hope \: it \: helps \: you ...... \\ Please \: mark \: my \: answer \: as \\ brainleist......$

Answer 2

Step-by-step explanation:

[(2^-1)+(4^-1)+(6^-1)+(8^-1)]^n = 1

=>1/2+1/4+1/6+1/8=(12+6+4+3)/24

=>25/24^0=1

there fore,n =0