The coefficient of x^4 in the expansion of (1 | vedclass

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

Question

(D) Given the expression $(1-x-x^2+x^3)^6$. We can factorize the expression as $[(1-x)-x^2(1-x)]^6 = [(1-x^2)(1-x)]^6 = (1-x^2)^6(1-x)^6$. Using the b

Vedclass · Accepted Answer

$-60$

Vedclass · Answer

(D) Given the expression $(1-x-x^2+x^3)^6$. We can factorize the expression as $[(1-x)-x^2(1-x)]^6 = [(1-x^2)(1-x)]^6 = (1-x^2)^6(1-x)^6$. Using the binomial expansion $(1+y)^n = \sum_{k=0}^{n} \binom{n}{k} y^k$: $(1-x^2)^6 = \binom{6}{0} - \binom{6}{1}x^2 + \binom{6}{2}x^4 - \dots = 1 - 6x^2 + 15x^4 - \dots$ $(1-x)^6 = \binom{6}{0} - \binom{6}{1}x + \binom{6}{2}x^2 - \binom{6}{3}x^3 + \binom{6}{4}x^4 - \dots = 1 - 6x + 15x^2 - 20x^3 + 15x^4 - \dots$ To find the coefficient of $x^4$ in the product $(1 - 6x^2 + 15x^4)(1 - 6x + 15x^2 - 20x^3 + 15x^4)$,we multiply terms that result in $x^4$: $1 	imes (15x^4) + (-6x^2) 	imes (15x^2) + (15x^4) 	imes (1) = 15 - 90 + 15 = -60$. Thus,the coefficient of $x^4$ is $-60$.

The coefficient of $x^4$ in the expansion of $(1-x-x^2+x^3)^6$ is

Similar Questions

Let $r$ be the remainder when $2021^{2020}$ is divided by $2020^2$. Then $r$ lies between

The coefficient of $x$ in the expansion of $[\sqrt{1 + x^2} - x]^{-1}$ in ascending powers of $x$,when $|x| < 1$,is

The last digit in $7^{300}$ is

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

The sum of the coefficients in the expansion of $(1+x+x^2)^n$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module