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Question

Physics Question on Mathematical Methods

Consider a binary number with m digits, where m is an even number. This binary number has alternating 1's and 0's, with digit 1 in the highest place value. The decimal equivalent of this binary number is

A

2m-1

B

(2m−1)3\frac{(2^m-1)}{3}

C

(2m+1−1)3\frac{(2^{m+1}-1)}{3}

D

23(2m−1)\frac{2}{3}(2^m-1)

Answer

23(2m−1)\frac{2}{3}(2^m-1)

Explanation

Solution

The correct answer is (D) : (2m+1−1)3\frac{(2^{m+1}-1)}{3}