Question

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. 2^m-1
• B. \(\frac{(2^m-1)}{3}\)
• C. \(\frac{(2^{m+1}-1)}{3}\)
• D. \(\frac{2}{3}(2^m-1)\)

Answer 1

The Correct Option is D

To solve this problem, we need to find the decimal equivalent of a binary number with \(m\) digits, where \(m\) is even. The given binary number alternates between 1's and 0's, with digit 1 at the highest place value.

Let's analyze how the binary number looks:

• With \(m\) digits, alternating 1's and 0's, starting with 1, the pattern of the binary digits will be: 1010...10. Since \(m\) is even, half of the digits will be 1's, and half will be 0's.

Let's calculate the decimal value of this binary number:

The binary number can be generalized as:

• If \(m = 2k\), for an integer \(k\), the binary sequence can be expressed as:
• 1 at positions: \(2k-1, 2k-3, \ldots, 1\)
• 0 at positions: \(2k-2, 2k-4, \ldots, 0\)

In decimal, this number is calculated by summing up the powers of 2 corresponding to each 1 in the binary sequence:

The decimal value \(D\) of this number can be expressed as:

• \(D = 2^{m-1} + 2^{m-3} + \cdots + 2^1\)
• This series is a geometric progression with the first term \(a = 2^1\) and common ratio \(r = 2^2 = 4\).

Using the formula for the sum of a geometric series \(S = a \frac{r^n - 1}{r - 1}\), where \(n\) is the number of terms, we can find the sum:

• The number of terms is \(n = k\) (since there are \(k\) ones).
• So, \(S = 2 \cdot \frac{(4^{k} - 1)}{3}\).

Substitute \(4 = 2^2\), and simplify the expression:

• \(S = 2 \cdot \frac{((2^2)^{k} - 1)}{3}\)
• \(S = 2 \cdot \frac{(2^{2k} - 1)}{3}\)

Notice that \(m = 2k\), so \(2k = m\), hence:

• \(S = 2 \cdot \frac{(2^{m} - 1)}{3}\)
• Thus, \(S = \frac{2}{3}(2^m - 1)\).