In the expansion of (1 + x)^{50},the sum of t | vedclass

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

Question

(B) We have $(1 + x)^{50} = \sum_{r=0}^{50} {}^{50}C_r x^r = {}^{50}C_0 + {}^{50}C_1 x + {}^{50}C_2 x^2 + {}^{50}C_3 x^3 + \dots + {}^{50}C_{50} x^{50

Vedclass · Accepted Answer

$2^{49}$

Vedclass · Answer

(B) We have $(1 + x)^{50} = \sum_{r=0}^{50} {}^{50}C_r x^r = {}^{50}C_0 + {}^{50}C_1 x + {}^{50}C_2 x^2 + {}^{50}C_3 x^3 + \dots + {}^{50}C_{50} x^{50}$.Let $S_o$ be the sum of coefficients of odd powers of $x$ and $S_e$ be the sum of coefficients of even powers of $x$.$S_o = {}^{50}C_1 + {}^{50}C_3 + \dots + {}^{50}C_{49}$ and $S_e = {}^{50}C_0 + {}^{50}C_2 + \dots + {}^{50}C_{50}$.We know that $(1 + 1)^{50} = S_e + S_o = 2^{50}$ and $(1 - 1)^{50} = S_e - S_o = 0$.Subtracting the two equations: $(S_e + S_o) - (S_e - S_o) = 2^{50} - 0$.$2S_o = 2^{50}$.$S_o = \frac{2^{50}}{2} = 2^{49}$.

In the expansion of $(1 + x)^{50}$,the sum of the coefficients of odd powers of $x$ is

Similar Questions

Let $n \in N$ and $[x]$ denote the greatest integer less than or equal to $x$. If the sum of $(n+1)$ terms ${}^{n}C_{0}, 3 \cdot {}^{n}C_{1}, 5 \cdot {}^{n}C_{2}, 7 \cdot {}^{n}C_{3}, \ldots$ is equal to $2^{100} \cdot 101$,then $2\left[\frac{n-1}{2}\right]$ is equal to $....$

The mean of the values $0, 1, 2, \dots, n$ having corresponding weights $^nC_0, ^nC_1, ^nC_2, \dots, ^nC_n$ respectively is

If $(1 + x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}$,then $a_0 + a_3 + a_6 + \dots =$

$\sum_{r=0}^{10} {}^{40-r} C_5$ is equal to

For $z \in \mathbb{C}$,if $(1+z)^n = 1 + { }^n C_1 z + { }^n C_2 z^2 + \ldots + { }^n C_n z^n$ and $\sum_{r=0}^{100} { }^{100} C_r \sin(rx) = \left(2 \cos \frac{x}{2}\right)^{100} \sin(kx)$,then $k =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module