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-

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) \dots (2 n - 2)}{1.3.5 \dots (n - 1)}

simplify to 2^{n - 1}

Answered by Smail last updated on 03/Apr/19

A = \frac{n (n + 2) (n + 4) \dots (2 n - 2)}{1.3.5 \dots (n - 1)}

n = 2 m

A = \frac{2 m (2 m + 2) (2 m + 4) \dots (4 m - 2)}{1.3.5 \dots (2 m - 1)}

= \frac{2 m . 2 (m + 1) 2 . (m + 2) \dots 2 (2 m - 1)}{1.3.5 \dots (2 m - 1)}

= \frac{2^{m} (2 m - 1)! \times (2.4.6 \dots (2 m - 2)}{(m - 1)! (1.2.3.4 \dots (2 m - 2) (2 m - 1)}

= \frac{2^{m} (2 m - 1)! \times 2^{m - 1} (1.2.3 \dots (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 \dots

Commented by Kunal12588 last updated on 03/Apr/19

t h a t w a s l i k e t h a t

\frac{2 m . 2 (m + 1) 2 . (m + 2) \dots 2 (2 m - 1)}{1.3.5 \dots (2 m - 1)}

= \frac{2^{m} m (m + 1) (m + 2) \dots (2 m - 1)}{1.3.5 \dots (2 m - 1)}

= \frac{2^{m}}{1.3.5 \dots (2 m - 1)} {m (m + 1) (m + 2) \dots (2 m - 1)}

\frac{2^{m}}{1.3.5 \dots (2 m - 1)} = k

∴ k {m (m + 1) (m + 2) \dots (2 m - 1)}

= k {\frac{1.2 \dots (m - 1) (m) (m + 1) \dots (2 m - 2) (2 m - 1)}{1.2.3 \dots (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) \dots (4 k - 2)}{1.3.5 \dots . (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) \dots . . (4 k + 2)}{1.3.5 \dots . (2 k + 1)} =

\frac{4 k (4 k + 2)}{2 k (2 k + 1)} \times \frac{2 k (2 k + 2) \dots (4 k - 2)}{1.3.5 \dots (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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