The sum, of the coefficients of the first 50 terms in the binomial expansion of $$(1 - x)^{100}$$, is equal to

The binomial expansion of $$(1-x)^{100}$$ is

$$(1-x)^{100}=\binom{100}{0}-\binom{100}{1}x+\binom{100}{2}x^2-\cdots+\binom{100}{100}x^{100}$$

The sum of the coefficients of the first 50 terms is

$$S=\binom{100}{0}-\binom{100}{1}+\binom{100}{2}-\cdots-\binom{100}{49}$$

The sum of all coefficients of the expansion can be obtained by putting $$x=1$$.

Therefore,

$$(1-1)^{100}=0$$

Hence,

$$\binom{100}{0}-\binom{100}{1}+\binom{100}{2}-\cdots+\binom{100}{100}=0$$

Since there are 101 terms, we split the expansion into:

• First 50 terms
• Middle term
• Last 50 terms

Thus,

$$S+\binom{100}{50}+S=0$$

Using the symmetry property $$\binom{n}{r}=\binom{n}{n-r}$$, the sum of the last 50 terms is equal to the sum of the first 50 terms.

Therefore,

$$2S+\binom{100}{50}=0$$

$$2S=-\binom{100}{50}$$

$$S=-\frac{1}{2}\binom{100}{50}$$

Now use the identity $$\binom{n}{r}=\frac{n}{r}\binom{n-1}{r-1}$$ so that

$$\binom{100}{50}=\frac{100}{50}\binom{99}{49}$$

$$\binom{100}{50}=2\binom{99}{49}$$

Substituting,

$$S=-\frac{1}{2}\left(2\binom{99}{49}\right)$$

$$S=-\binom{99}{49}$$

Hence, the sum of the coefficients of the first 50 terms is

$$-\binom{99}{49}$$

Therefore, the correct answer is

$$-\binom{99}{49}$$