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Question

(C) The binomial expansion is $(1-x)^{100} = C_0 - C_1x + C_2x^2 - C_3x^3 + \dots + C_{100}x^{100}$.Let $S = C_0 - C_1 + C_2 - C_3 + \dots - C_{49}$.W

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The binomial expansion is $(1-x)^{100} = C_0 - C_1x + C_2x^2 - C_3x^3 + \dots + C_{100}x^{100}$.Let $S = C_0 - C_1 + C_2 - C_3 + \dots - C_{49}$.We know that the sum of all coefficients in $(1-x)^{100}$ is $(1-1)^{100} = 0$.Thus,$(C_0 - C_1 + C_2 - \dots + C_{50} - \dots + C_{100}) = 0$.Using the property $C_r = C_{n-r}$,we have $C_{100} = C_0, C_{99} = C_1, \dots, C_{51} = C_{49}$.So,$2(C_0 - C_1 + C_2 - \dots - C_{49}) + C_{50} = 0$.$2S + C_{50} = 0 \implies S = -\frac{1}{2} C_{50}$.$S = -\frac{1}{2} \binom{100}{50} = -\frac{1}{2} 	imes \frac{100}{50} \binom{99}{49} = -\binom{99}{49}$.

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

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