If n be even, show that the expression ((n(n + 2)(n + 4) ... (2n − 2))/(1.3.5 ... (n − 1))) simplify to 2^(n − 1)


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Question Number 57336 by Tawa1 last updated on 02/Apr/19

$If n be even, show that the expression \frac{n (n + 2) (n + 4) . . . (2 n - 2)}{1.3.5 . . . (n - 1)}$ $simplify to 2^{n - 1}$
	Answered by Smail last updated on 03/Apr/19

	$A = \frac{n (n + 2) (n + 4) . . . (2 n - 2)}{1.3.5 . . . (n - 1)}$ $n = 2 m$ $A = \frac{2 m (2 m + 2) (2 m + 4) . . . (4 m - 2)}{1.3.5 . . . (2 m - 1)}$ $= \frac{2 m . 2 (m + 1) 2 . (m + 2) . . . 2 (2 m - 1)}{1.3.5 . . . (2 m - 1)}$ $= \frac{2^{m} (2 m - 1)! \times (2.4.6 . . . (2 m - 2)}{(m - 1)! (1.2.3.4 . . . (2 m - 2) (2 m - 1)}$ $= \frac{2^{m} (2 m - 1)! \times 2^{m - 1} (1.2.3 . . . (m - 1))}{(m - 1)! \times (2 m - 1)!}$ $= \frac{2^{m} \times 2^{m - 1} (2 m - 1)! \times (m - 1)!}{(m - 1)! \times (2 m - 1)!}$ $= 2^{2 m - 1} = 2^{n - 1}$
	Commented by Tawa1 last updated on 03/Apr/19

	$God bless you sir .$ $sir please am confused from where you started using the blue$ $ink . I want to understand sir . Thanks for your time .$
	Commented by Tawa1 last updated on 03/Apr/19

	$How does everything becomes (2 m - 1)!$ $and later 2^{m - 1} (1.2.3 . . .$
	Commented by Kunal12588 last updated on 03/Apr/19
	$that was like that ((2m.2(m+1)2.(m+2)...2(2m−1))/(1.3.5...(2m−1))) =((2^m m(m+1)(m+2)...(2m−1))/(1.3.5...(2m−1))) =(2^m /(1.3.5...(2m−1))){m(m+1)(m+2)...(2m−1)} (2^m /(1.3.5...(2m−1)))=k ∴k{m(m+1)(m+2)...(2m−1)} =k{((1.2...(m−1)(m)(m+1)...(2m−2)(2m−1))/(1.2.3...(m−1)))} =k(((2m−1)!)/((m−1)!))$
	$t h a t w a s l i k e t h a t$ $\frac{2 m . 2 (m + 1) 2 . (m + 2) . . . 2 (2 m - 1)}{1.3.5 . . . (2 m - 1)}$ $= \frac{2^{m} m (m + 1) (m + 2) . . . (2 m - 1)}{1.3.5 . . . (2 m - 1)}$ $= \frac{2^{m}}{1.3.5 . . . (2 m - 1)} {m (m + 1) (m + 2) . . . (2 m - 1)}$ $\frac{2^{m}}{1.3.5 . . . (2 m - 1)} = k$ $∴ k {m (m + 1) (m + 2) . . . (2 m - 1)}$ $= k {\frac{1.2 . . . (m - 1) (m) (m + 1) . . . (2 m - 2) (2 m - 1)}{1.2.3 . . . (m - 1)}}$ $= k \frac{(2 m - 1)!}{(m - 1)!}$
	Commented by Tawa1 last updated on 03/Apr/19

	$I appreciate sir . God bless you$
	Answered by kaivan.ahmadi last updated on 02/Apr/19

	$b y i n d u c t i o n i f n = 2 \Rightarrow \frac{2}{1} = 2^{2 - 1}$ $i n d u c t i o n a s s u m p t i o n$ $i f n = 2 k \Rightarrow \frac{2 k (2 k + 2) . . . (4 k - 2)}{1.3.5 . . . . (2 k - 1)} = 2^{2 k - 1}$ $n o w l e t n = 2 k + 2 \Rightarrow$ $\frac{(2 k + 2) (2 k + 4) . . . . . (4 k + 2)}{1.3.5 . . . . (2 k + 1)} =$ $\frac{4 k (4 k + 2)}{2 k (2 k + 1)} \times \frac{2 k (2 k + 2) . . . (4 k - 2)}{1.3.5 . . . (2 k - 1)} =$ $2 \times 2 \times 2^{2 k - 1} = 2^{2 k + 1}$
	Commented by Tawa1 last updated on 03/Apr/19

	$God bless you sir$
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