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

(B) The sum of the coefficients of a polynomial $P(x)$ is obtained by setting all variables equal to $1$.For the first expansion $(\alpha x^2 - 2x + 1

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) The sum of the coefficients of a polynomial $P(x)$ is obtained by setting all variables equal to $1$.For the first expansion $(\alpha x^2 - 2x + 1)^{35}$,the sum of coefficients is $(\alpha(1)^2 - 2(1) + 1)^{35} = (\alpha - 2 + 1)^{35} = (\alpha - 1)^{35}$.For the second expansion $(x - \alpha y)^{35}$,the sum of coefficients is $(1 - \alpha(1))^{35} = (1 - \alpha)^{35}$.Given that these sums are equal: $(\alpha - 1)^{35} = (1 - \alpha)^{35}$.Since $35$ is an odd integer,$(1 - \alpha)^{35} = -(\alpha - 1)^{35}$.Thus,$(\alpha - 1)^{35} = -(\alpha - 1)^{35}$,which implies $2(\alpha - 1)^{35} = 0$.Therefore,$(\alpha - 1)^{35} = 0$,which gives $\alpha - 1 = 0$,so $\alpha = 1$.

If the sum of the coefficients in the expansion of $(\alpha x^2 - 2x + 1)^{35}$ is equal to the sum of the coefficients in the expansion of $(x - \alpha y)^{35}$,then $\alpha = $

Similar Questions

The value of $(\sqrt{5} + 1)^5 - (\sqrt{5} - 1)^5$ is

Using the Binomial Theorem,evaluate $(96)^{3}$.

Let $n \geq 5$ be an integer. If $9^{n}-8n-1=64\alpha$ and $6^{n}-5n-1=25\beta$,then $\alpha-\beta$ is equal to

The total number of terms in the expansion of $(x+a)^{47}-(x-a)^{47}$ after simplification is

Find $(a+b)^{4}-(a-b)^{4}$. Hence,evaluate $(\sqrt{3}+\sqrt{2})^{4}-(\sqrt{3}-\sqrt{2})^{4}$. (in $\sqrt{6}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module