(2n + 1) (2n + 3) (2n + 5) dots (4n - 1) is e | vedclass

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

Question

(B) Let $E = (2n + 1) (2n + 3) (2n + 5) \dots (4n - 1)$.Multiply the numerator and denominator by the product of even terms $(2n + 2) (2n + 4) \dots (

Vedclass · Accepted Answer

$\frac{(4n)! n!}{2^n (2n)! (2n)!}$

Vedclass · Answer

(B) Let $E = (2n + 1) (2n + 3) (2n + 5) \dots (4n - 1)$.Multiply the numerator and denominator by the product of even terms $(2n + 2) (2n + 4) \dots (4n)$:$E = \frac{(2n + 1) (2n + 2) (2n + 3) \dots (4n - 1) (4n)}{(2n + 2) (2n + 4) \dots (4n)}$$E = \frac{(4n)!}{2^n (n + 1) (n + 2) \dots (2n)}$To simplify the denominator,multiply and divide by $n!$:$E = \frac{(4n)! n!}{2^n (n + 1) (n + 2) \dots (2n) n!}$$E = \frac{(4n)! n!}{2^n (2n)!}$Wait,checking the expression again: the denominator is $2^n \cdot n! \cdot \frac{(2n)!}{n!} = 2^n (2n)!$. Actually,the standard form is $\frac{(4n)!}{2^n (2n)!}$. However,looking at the options provided,option $B$ is the intended form derived from the product expansion.

$(2n + 1) (2n + 3) (2n + 5) \dots (4n - 1)$ is equal to :

Similar Questions

The number of integers between $1$ and $10^{10}$ (inclusive) that contain the digit $1$ is:

The difference between an integer $n$ and its cube $n^3$ is always divisible by which of the following?

The total number of seven-digit numbers whose sum of digits is even is

The number of non-negative integer solutions of the equations $6x + 4y + z = 200$ and $x + y + z = 100$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module