If the sum of the coefficients in the expansi | vedclass

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

Question

(A) The sum of the coefficients in the expansion of a polynomial $P(x, y, z)$ is found by setting all variables equal to $1$.For the expansion $(x - 2

Vedclass · Accepted Answer

$35$

Vedclass · Answer

(A) The sum of the coefficients in the expansion of a polynomial $P(x, y, z)$ is found by setting all variables equal to $1$.For the expansion $(x - 2y + 3z)^n$,the sum of coefficients is $(1 - 2 + 3)^n = 2^n$.Given $2^n = 128$,we have $2^n = 2^7$,which implies $n = 7$.The expansion of $(1 + x)^7$ is given by $\sum_{k=0}^{7} {^7C_k} x^k$.The coefficients are ${^7C_0}, {^7C_1}, {^7C_2}, {^7C_3}, {^7C_4}, {^7C_5}, {^7C_6}, {^7C_7}$.The greatest coefficient is the middle term,which is ${^7C_3}$ or ${^7C_4}$.Calculating the value: ${^7C_3} = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35$.

If the sum of the coefficients in the expansion of $(x - 2y + 3z)^n$,$n \in N$ is $128$,then the greatest coefficient in the expansion of $(1 + x)^n$ is

Similar Questions

If the coefficient of $x^3$ in the binomial expansion of $x^3(2 \sqrt{3} x^2 + \frac{1}{kx})^{12}$ is $880$,then $k$ is equal to

The $13^{th}$ term in the expansion of $\left(x^{2}+\frac{2}{x}\right)^{n}$ is independent of $x$. Then the sum of the divisors of $n$ is:

The term independent of $x$ in the expansion of $\left( 9x - \frac{1}{3\sqrt{x}} \right)^{18}, x > 0$,is $\alpha$ times the corresponding binomial coefficient. Then $\alpha$ is:

If the coefficients of $r$ th and $(r+1)$ th terms in the expansion of $(3+7x)^{29}$ are equal,then $r$ is equal to

The square root of the independent term in the expansion of $\left(\frac{2x^2}{5} + \sqrt{\frac{5}{x}}\right)^{10}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module