JEE Mathematics 2023 Question with Solution

Given: We need the sum of coefficients of the first $50$ terms in the expansion of $$ (1-x)^{100}

$$$$

Find: The required sum and the correct option.

Using the binomial expansion,

$$(1-x)^{100} = {}^{100}C_0 - {}^{100}C_1 x + {}^{100}C_2 x^2 - {}^{100}C_3 x^3 + \cdots - {}^{100}C_{99} x^{99} + {}^{100}C_{100} x^{100}$$

So the sum of all coefficients is obtained by putting $x=1$:

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

Hence,

$${}^{100}C_0 - {}^{100}C_1 + {}^{100}C_2 - {}^{100}C_3 + \cdots - {}^{100}C_{99} + {}^{100}C_{100} = 0$$

By symmetry of binomial coefficients, pairing terms gives

$$2\left({}^{100}C_0 - {}^{100}C_1 + {}^{100}C_2 - \cdots - {}^{100}C_{49}\right) + {}^{100}C_{50} = 0$$

Therefore,

$${}^{100}C_0 - {}^{100}C_1 + {}^{100}C_2 - \cdots - {}^{100}C_{49} = -\frac{1}{2}{}^{100}C_{50}$$

Now use the identity

$${}^{100}C_{50} = 2\,{}^{99}C_{49}$$

So,

$$-\frac{1}{2}{}^{100}C_{50} = -{}^{99}C_{49}$$

Therefore, the required sum is $-{}^{99}C_{49}$. The solution working gives the correct value, which corresponds numerically to option C, although the solution incorrectly labels it as B.

From

$${}^{100}C_0 - {}^{100}C_1 + {}^{100}C_2 - \cdots + {}^{100}C_{100} = 0$$

and the symmetry

$${}^{100}C_r = {}^{100}C_{100-r}$$

each pair from the beginning and end contributes equally. Since $100$ is even, the middle term is

$${}^{100}C_{50}$$

Thus,

$$2S + {}^{100}C_{50} = 0$$

where

$$S = {}^{100}C_0 - {}^{100}C_1 + {}^{100}C_2 - \cdots - {}^{100}C_{49}$$

Hence,

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

Now,

$${}^{100}C_{50} = \frac{100!}{50!50!}$$

and

$${}^{99}C_{49} = \frac{99!}{49!50!}$$

Therefore,

$${}^{100}C_{50} = 2\,{}^{99}C_{49}$$

which gives

$$S = -{}^{99}C_{49}$$

So the correct option is C.