The constant term in the expansion of left(2x | vedclass

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

Question

(B) The general term in the multinomial expansion is given by $\frac{5!}{n_1! n_2! n_3!} (2x)^{n_1} (x^{-7})^{n_2} (3x^2)^{n_3}$,where $n_1 + n_2 + n_

Vedclass · Accepted Answer

$1080$

Vedclass · Answer

(B) The general term in the multinomial expansion is given by $\frac{5!}{n_1! n_2! n_3!} (2x)^{n_1} (x^{-7})^{n_2} (3x^2)^{n_3}$,where $n_1 + n_2 + n_3 = 5$.This simplifies to $\frac{5!}{n_1! n_2! n_3!} 2^{n_1} 3^{n_3} x^{n_1 - 7n_2 + 2n_3}$.For the constant term,the power of $x$ must be zero,so $n_1 - 7n_2 + 2n_3 = 0$.Given $n_1 + n_2 + n_3 = 5$,we substitute $n_1 = 5 - n_2 - n_3$ into the equation:$(5 - n_2 - n_3) - 7n_2 + 2n_3 = 0$ $\Rightarrow 5 - 8n_2 + n_3 = 0$ $\Rightarrow n_3 = 8n_2 - 5$.If $n_2 = 1$,then $n_3 = 3$,which implies $n_1 = 5 - 1 - 3 = 1$.Substituting these values into the coefficient formula:$\frac{5!}{1! 1! 3!} (2)^1 (3)^3 = \frac{120}{6} 	imes 2 	imes 27 = 20 	imes 54 = 1080$.

The constant term in the expansion of $\left(2x + \frac{1}{x^7} + 3x^2\right)^5$ is $........$.

Similar Questions

The coefficient of $x^5$ in $(3+x+x^2)^6$ is

The total number of $4$-digit numbers whose greatest common divisor with $18$ is $3$ is .... .

Let $N = n(n+1)(n+2)(n+3)$ where $n$ is a natural number,and $d$ is the number of divisors of $N$. Which of the following is true?

The number of positive divisors of $1080$ is

The largest natural number $n$ such that $3^{n}$ divides $66!$ is $............$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module