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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) We have the expression $E = {2^n} \{ 1 \cdot 3 \cdot 5 \cdots (2n - 1) \}$.To simplify this,multiply and divide by the product of even numbers $2 \cdot 4 \cdot 6 \cdots (2n)$:$E = \frac{{2^n} \{ 1 \cdot 3 \cdot 5 \cdots (2n - 1) \} \cdot \{ 2 \cdot 4 \cdot 6 \cdots (2n) \}}{2 \cdot 4 \cdot 6 \cdots (2n)}$$E = \frac{(2n)!}{2 \cdot 1 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdots 2 \cdot n}$$E = \frac{(2n)!}{{2^n} \cdot (1 \cdot 2 \cdot 3 \cdots n)}$$E = \frac{(2n)!}{{2^n} \cdot n!} \cdot {2^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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