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Maths-

General

Easy

Question

$left parenthesis 2 to the power of 3 n end exponent minus 1 right parenthesis$ will be divisible by $left parenthesis for all n element of N right parenthesis$

25
8
7
3

The correct answer is: 7

$2 to the power of 3 n end exponent minus 1 equals left parenthesis 2 to the power of 3 end exponent right parenthesis to the power of n end exponent minus 1$
$equals 8 to the power of n end exponent minus 1 equals left parenthesis 1 plus 7 right parenthesis to the power of n end exponent minus 1$
$equals 1 plus blank to the power of n end exponent C subscript 1 end subscript 7 plus blank to the power of n end exponent C subscript 2 end subscript 7 to the power of 2 end exponent plus$ … $plus blank to the power of n end exponent C subscript n end subscript 7 to the power of n end exponent minus 1$
$equals 7 left square bracket blank to the power of n end exponent C subscript 1 end subscript plus blank to the power of n end exponent C subscript 2 end subscript 7 plus$ … $plus blank to the power of n end exponent C subscript n end subscript 7 to the power of n minus 1 end exponent right square bracket$
$therefore 2 to the power of 3 n end exponent minus 1$ is divisible by 7

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