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

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)

$$\mathrm{If}\:\:\mathrm{n}\:\mathrm{be}\:\mathrm{even},\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{expression}\:\:\:\:\frac{\mathrm{n}\left(\mathrm{n}\:+\:\mathrm{2}\right)\left(\mathrm{n}\:+\:\mathrm{4}\right)\:...\:\left(\mathrm{2n}\:−\:\mathrm{2}\right)}{\mathrm{1}.\mathrm{3}.\mathrm{5}\:...\:\left(\mathrm{n}\:−\:\mathrm{1}\right)} \\ $$$$\mathrm{simplify}\:\mathrm{to}\:\:\mathrm{2}^{\mathrm{n}\:−\:\mathrm{1}} \\ $$

Answered by Smail last updated on 03/Apr/19

A=((n(n+2)(n+4)...(2n−2))/(1.3.5...(n−1))) n=2m A=((2m(2m+2)(2m+4)...(4m−2))/(1.3.5...(2m−1))) =((2m.2(m+1)2.(m+2)...2(2m−1))/(1.3.5...(2m−1))) =((2^m (2m−1)!×(2.4.6...(2m−2))/((m−1)!(1.2.3.4...(2m−2)(2m−1))) =((2^m (2m−1)!×2^(m−1) (1.2.3...(m−1)))/((m−1)!×(2m−1)!)) =((2^m ×2^(m−1) (2m−1)!×(m−1)!)/((m−1)!×(2m−1)!)) =2^(2m−1) =2^(n−1)

$${A}=\frac{{n}\left({n}+\mathrm{2}\right)\left({n}+\mathrm{4}\right)...\left(\mathrm{2}{n}−\mathrm{2}\right)}{\mathrm{1}.\mathrm{3}.\mathrm{5}...\left({n}−\mathrm{1}\right)} \\ $$$${n}=\mathrm{2}{m} \\ $$$${A}=\frac{\mathrm{2}{m}\left(\mathrm{2}{m}+\mathrm{2}\right)\left(\mathrm{2}{m}+\mathrm{4}\right)...\left(\mathrm{4}{m}−\mathrm{2}\right)}{\mathrm{1}.\mathrm{3}.\mathrm{5}...\left(\mathrm{2}{m}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{2}{m}.\mathrm{2}\left({m}+\mathrm{1}\right)\mathrm{2}.\left({m}+\mathrm{2}\right)...\mathrm{2}\left(\mathrm{2}{m}−\mathrm{1}\right)}{\mathrm{1}.\mathrm{3}.\mathrm{5}...\left(\mathrm{2}{m}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{2}^{{m}} \left(\mathrm{2}{m}−\mathrm{1}\right)!×\left(\mathrm{2}.\mathrm{4}.\mathrm{6}...\left(\mathrm{2}{m}−\mathrm{2}\right)\right.}{\left({m}−\mathrm{1}\right)!\left(\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}...\left(\mathrm{2}{m}−\mathrm{2}\right)\left(\mathrm{2}{m}−\mathrm{1}\right)\right.} \\ $$$$=\frac{\mathrm{2}^{{m}} \left(\mathrm{2}{m}−\mathrm{1}\right)!×\mathrm{2}^{{m}−\mathrm{1}} \left(\mathrm{1}.\mathrm{2}.\mathrm{3}...\left({m}−\mathrm{1}\right)\right)}{\left({m}−\mathrm{1}\right)!×\left(\mathrm{2}{m}−\mathrm{1}\right)!} \\ $$$$=\frac{\mathrm{2}^{{m}} ×\mathrm{2}^{{m}−\mathrm{1}} \left(\mathrm{2}{m}−\mathrm{1}\right)!×\left({m}−\mathrm{1}\right)!}{\left({m}−\mathrm{1}\right)!×\left(\mathrm{2}{m}−\mathrm{1}\right)!} \\ $$$$=\mathrm{2}^{\mathrm{2}{m}−\mathrm{1}} =\mathrm{2}^{{n}−\mathrm{1}} \\ $$

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.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{sir}\:\mathrm{please}\:\mathrm{am}\:\mathrm{confused}\:\mathrm{from}\:\mathrm{where}\:\mathrm{you}\:\mathrm{started}\:\mathrm{using}\:\mathrm{the}\:\mathrm{blue} \\ $$$$\mathrm{ink}.\:\:\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{sir}.\:\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}. \\ $$

Commented by Tawa1 last updated on 03/Apr/19

How does everything becomes (2m − 1)! and later 2^(m − 1) (1.2.3 ...

$$\mathrm{How}\:\mathrm{does}\:\mathrm{everything}\:\mathrm{becomes}\:\:\left(\mathrm{2m}\:−\:\mathrm{1}\right)! \\ $$$$\mathrm{and}\:\mathrm{later}\:\:\:\mathrm{2}^{\mathrm{m}\:−\:\mathrm{1}} \:\left(\mathrm{1}.\mathrm{2}.\mathrm{3}\:...\right. \\ $$

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)!))

$${that}\:{was}\:{like}\:{that} \\ $$$$\frac{\mathrm{2}{m}.\mathrm{2}\left({m}+\mathrm{1}\right)\mathrm{2}.\left({m}+\mathrm{2}\right)...\mathrm{2}\left(\mathrm{2}{m}−\mathrm{1}\right)}{\mathrm{1}.\mathrm{3}.\mathrm{5}...\left(\mathrm{2}{m}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{2}^{{m}} {m}\left({m}+\mathrm{1}\right)\left({m}+\mathrm{2}\right)...\left(\mathrm{2}{m}−\mathrm{1}\right)}{\mathrm{1}.\mathrm{3}.\mathrm{5}...\left(\mathrm{2}{m}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{2}^{{m}} }{\mathrm{1}.\mathrm{3}.\mathrm{5}...\left(\mathrm{2}{m}−\mathrm{1}\right)}\left\{{m}\left({m}+\mathrm{1}\right)\left({m}+\mathrm{2}\right)...\left(\mathrm{2}{m}−\mathrm{1}\right)\right\} \\ $$$$\frac{\mathrm{2}^{{m}} }{\mathrm{1}.\mathrm{3}.\mathrm{5}...\left(\mathrm{2}{m}−\mathrm{1}\right)}={k} \\ $$$$\therefore{k}\left\{{m}\left({m}+\mathrm{1}\right)\left({m}+\mathrm{2}\right)...\left(\mathrm{2}{m}−\mathrm{1}\right)\right\} \\ $$$$={k}\left\{\frac{\mathrm{1}.\mathrm{2}...\left({m}−\mathrm{1}\right)\left({m}\right)\left({m}+\mathrm{1}\right)...\left(\mathrm{2}{m}−\mathrm{2}\right)\left(\mathrm{2}{m}−\mathrm{1}\right)}{\mathrm{1}.\mathrm{2}.\mathrm{3}...\left({m}−\mathrm{1}\right)}\right\} \\ $$$$={k}\frac{\left(\mathrm{2}{m}−\mathrm{1}\right)!}{\left({m}−\mathrm{1}\right)!} \\ $$

Commented by Tawa1 last updated on 03/Apr/19

I appreciate sir. God bless you

$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$

Answered by kaivan.ahmadi last updated on 02/Apr/19

by induction if n=2⇒(2/1)=2^(2−1) induction assumption if n=2k⇒((2k(2k+2)...(4k−2))/(1.3.5....(2k−1)))=2^(2k−1) now let n=2k+2⇒ (((2k+2)(2k+4).....(4k+2))/(1.3.5....(2k+1)))= ((4k(4k+2))/(2k(2k+1)))×((2k(2k+2)...(4k−2))/(1.3.5...(2k−1)))= 2×2×2^(2k−1) =2^(2k+1)

$${by}\:{induction}\:{if}\:{n}=\mathrm{2}\Rightarrow\frac{\mathrm{2}}{\mathrm{1}}=\mathrm{2}^{\mathrm{2}−\mathrm{1}} \\ $$$${induction}\:{assumption} \\ $$$${if}\:{n}=\mathrm{2}{k}\Rightarrow\frac{\mathrm{2}{k}\left(\mathrm{2}{k}+\mathrm{2}\right)...\left(\mathrm{4}{k}−\mathrm{2}\right)}{\mathrm{1}.\mathrm{3}.\mathrm{5}....\left(\mathrm{2}{k}−\mathrm{1}\right)}=\mathrm{2}^{\mathrm{2}{k}−\mathrm{1}} \\ $$$${now}\:{let}\:{n}=\mathrm{2}{k}+\mathrm{2}\Rightarrow \\ $$$$\frac{\left(\mathrm{2}{k}+\mathrm{2}\right)\left(\mathrm{2}{k}+\mathrm{4}\right).....\left(\mathrm{4}{k}+\mathrm{2}\right)}{\mathrm{1}.\mathrm{3}.\mathrm{5}....\left(\mathrm{2}{k}+\mathrm{1}\right)}= \\ $$$$\frac{\mathrm{4}{k}\left(\mathrm{4}{k}+\mathrm{2}\right)}{\mathrm{2}{k}\left(\mathrm{2}{k}+\mathrm{1}\right)}×\frac{\mathrm{2}{k}\left(\mathrm{2}{k}+\mathrm{2}\right)...\left(\mathrm{4}{k}−\mathrm{2}\right)}{\mathrm{1}.\mathrm{3}.\mathrm{5}...\left(\mathrm{2}{k}−\mathrm{1}\right)}= \\ $$$$\mathrm{2}×\mathrm{2}×\mathrm{2}^{\mathrm{2}{k}−\mathrm{1}} =\mathrm{2}^{\mathrm{2}{k}+\mathrm{1}} \\ $$

Commented by Tawa1 last updated on 03/Apr/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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