The sum of the coefficients of the first 50 t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

The greatest coefficient in the expansion of $(1 + x)^{2n + 2}$ is

If $L$ and $M$ are respectively the coefficient of $x^{-7}$ in $\left(a x+\frac{b}{x^2}\right)^{11}$ and the coefficient of $x^7$ in $\left(b x^2+\frac{a}{x}\right)^{11}$,then $L+M=$

The coefficient of $x^{10}$ in the expansion of $1+(1+x)+\dots+(1+x)^{20}$

The coefficient of the term independent of $x$ in the expansion of $(1 + x + 2x^3)\left( \frac{3}{2}x^2 - \frac{1}{3x} \right)^9$ is

In the binomial expansion of $(1+x)^{15}$,the coefficients of $x^{r}$ and $x^{r+3}$ are equal. Then,$r$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module