The value of {2^n} { 1 cdot 3 cdot 5 cdots (2 | vedclass

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Question

(A) We are given the expression $E = 2^n \{ 1 \cdot 3 \cdot 5 \cdots (2n - 1) \}$.To simplify this,multiply and divide by the product of the first $n$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) We are given the expression $E = 2^n \{ 1 \cdot 3 \cdot 5 \cdots (2n - 1) \}$.To simplify this,multiply and divide by the product of the first $n$ even numbers,which is $2 \cdot 4 \cdot 6 \cdots (2n) = 2^n \cdot (1 \cdot 2 \cdot 3 \cdots n) = 2^n \cdot n!$.Thus,$E = \frac{2^n \cdot \{ 1 \cdot 3 \cdot 5 \cdots (2n - 1) \} \cdot \{ 2 \cdot 4 \cdot 6 \cdots (2n) \}}{2 \cdot 4 \cdot 6 \cdots (2n)}$.The numerator becomes the product of all integers from $1$ to $2n$,which is $(2n)!$.The denominator is $2^n \cdot n!$.Therefore,$E = \frac{(2n)! \cdot 2^n}{2^n \cdot n!} = \frac{(2n)!}{n!}$.

The value of ${2^n} \{ 1 \cdot 3 \cdot 5 \cdots (2n - 3) \cdot (2n - 1) \}$ is

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