To simplify the expression \( B = 2^{50} - 2^{49} + 2^{48} - 2^{47} + \cdots + 2^4 - 2^3 \), we can factor out the common term \( 2^3 \) from each power of 2 in the series. Let's rewrite the expression:

\[ B = 2^{50} - 2^{49} + 2^{48} - 2^{47} + \cdots + 2^4 - 2^3 \]

Notice that each term in the series can be written as \( 2^k \) where \( k \) ranges from 3 to 50. We can factor out \( 2^3 \) from each term:

\[ B = 2^3 (2^{47} - 2^{46} + 2^{45} - 2^{44} + \cdots + 2^1 - 2^0) \]

Now, let's simplify the expression inside the parentheses:

\[ 2^{47} - 2^{46} + 2^{45} - 2^{44} + \cdots + 2^1 - 2^0 \]

This is a geometric series with the first term \( a = 2^{47} \) and the common ratio \( r = -\frac{1}{2} \). The number of terms in the series is 48 (from \( 2^{47} \) to \( 2^0 \)).

The sum \( S \) of a geometric series with \( n \) terms, first term \( a \), and common ratio \( r \) is given by:

\[ S = a \frac{1 - r^n}{1 - r} \]

In this case, \( a = 2^{47} \), \( r = -\frac{1}{2} \), and \( n = 48 \):

\[ S = 2^{47} \frac{1 - \left(-\frac{1}{2}\right)^{48}}{1 - \left(-\frac{1}{2}\right)} \]

Since \( \left(-\frac{1}{2}\right)^{48} \) is a very small number (essentially approaching 0), we can approximate it as 0 for simplicity:

\[ S \approx 2^{47} \frac{1 - 0}{1 - \left(-\frac{1}{2}\right)} = 2^{47} \frac{1}{1 + \frac{1}{2}} = 2^{47} \frac{1}{\frac{3}{2}} = 2^{47} \cdot \frac{2}{3} = \frac{2^{48}}{3} \]

Therefore, the sum inside the parentheses is approximately \( \frac{2^{48}}{3} \).

Now, multiply this result by \( 2^3 \):

\[ B = 2^3 \cdot \frac{2^{48}}{3} = \frac{2^{51}}{3} \]

So, the simplified form of \( B \) is:

\[ B = \frac{2^{51}}{3} \]