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Question Number 72965 by Raxreedoroid last updated on 05/Nov/19
Prove that this equation is true: Π_(x=1) ^(n−1) (2x+1)=(((2x−1)!)/((2)^(x−1) (x−1)!))
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{this}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{true}: \\ $$$$\underset{{x}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{2}{x}+\mathrm{1}\right)=\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)!}{\left(\mathrm{2}\right)^{{x}−\mathrm{1}} \left({x}−\mathrm{1}\right)!} \\ $$
Commented by mathmax by abdo last updated on 05/Nov/19
let prove by recurrence that Π_(k=1) ^(n−1) (2k+1)=(((2n−1)!)/(2^(n−1) (n−1)!)) (α) n=2 ⇒3=((3!)/2) =3( relation true) let suppose(α)true ⇒ Π_(k=1) ^n (2k+1) =Π_(k=1) ^(n−1) (2k+1)(2n+1) =(((2n−1)!)/(2^(n−1) (n−1)!))(2n+1) =(((2n+1)(2n)(2n−1)!)/((2n)2^(n−1) (n−1)!)) =(((2n+1)!)/(2^n n!)) the relation is true at termn+1. but there is a error in the question the correct is Π_(x=1) ^(n−1) (2x+1)=(((2n−1)!)/(2^(n−1) (n−1)!))
$${let}\:{prove}\:{by}\:{recurrence}\:\:{that}\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(\mathrm{2}{k}+\mathrm{1}\right)=\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!}{\mathrm{2}^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}\:\:\left(\alpha\right) \\ $$$${n}=\mathrm{2}\:\Rightarrow\mathrm{3}=\frac{\mathrm{3}!}{\mathrm{2}}\:=\mathrm{3}\left(\:{relation}\:{true}\right)\:\:{let}\:{suppose}\left(\alpha\right){true}\:\Rightarrow \\ $$$$\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{2}{k}+\mathrm{1}\right)\:=\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\:=\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!}{\mathrm{2}^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$$=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)!}{\left(\mathrm{2}{n}\right)\mathrm{2}^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}\:=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{{n}} \:{n}!}\:\:\:\:{the}\:{relation}\:{is}\:{true}\:{at}\:{termn}+\mathrm{1}. \\ $$$${but}\:{there}\:{is}\:{a}\:{error}\:{in}\:{the}\:{question}\:{the}\:{correct}\:{is} \\ $$$$\prod_{{x}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(\mathrm{2}{x}+\mathrm{1}\right)=\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!}{\mathrm{2}^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!} \\ $$
Answered by mind is power last updated on 05/Nov/19
(((2n−1)...3.1.(2n−2)......2)/((2n−2)......2.))=(((2n−1)!)/(2(n−1).2(n−2)....2.1))=(((2n−1)!)/(2^(n−1) .(n−1)!)) ⇒Π_(k=1) ^(n−1) (2k+1)=(((2n−1)!)/(2^(n−1) (n−1)!))
$$\frac{\left(\mathrm{2n}−\mathrm{1}\right)…\mathrm{3}.\mathrm{1}.\left(\mathrm{2n}−\mathrm{2}\right)……\mathrm{2}}{\left(\mathrm{2n}−\mathrm{2}\right)……\mathrm{2}.}=\frac{\left(\mathrm{2n}−\mathrm{1}\right)!}{\mathrm{2}\left(\mathrm{n}−\mathrm{1}\right).\mathrm{2}\left(\mathrm{n}−\mathrm{2}\right)….\mathrm{2}.\mathrm{1}}=\frac{\left(\mathrm{2n}−\mathrm{1}\right)!}{\mathrm{2}^{\mathrm{n}−\mathrm{1}} .\left(\mathrm{n}−\mathrm{1}\right)!} \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\prod}}\left(\mathrm{2k}+\mathrm{1}\right)=\frac{\left(\mathrm{2n}−\mathrm{1}\right)!}{\mathrm{2}^{\mathrm{n}−\mathrm{1}} \left(\mathrm{n}−\mathrm{1}\right)!} \\ $$
Commented by Raxreedoroid last updated on 05/Nov/19
Your answer isn′t clear to me why did you put ′′3.1.′′ and ′′2′′s there
$$\mathrm{Your}\:\mathrm{answer}\:\mathrm{isn}'\mathrm{t}\:\mathrm{clear}\:\mathrm{to}\:\mathrm{me}\: \\ $$$$\mathrm{why}\:\mathrm{did}\:\mathrm{you}\:\mathrm{put}\:''\mathrm{3}.\mathrm{1}.''\:\mathrm{and}\:''\mathrm{2}''\mathrm{s}\:\mathrm{there} \\ $$
Commented by ajfour last updated on 05/Nov/19
whats the last digit of 1!+2!+3!+.....+99!
$${whats}\:{the}\:{last}\:{digit}\:{of} \\ $$$$\mathrm{1}!+\mathrm{2}!+\mathrm{3}!+…..+\mathrm{99}!\: \\ $$
Commented by Henri Boucatchou last updated on 05/Nov/19
Check once more brother; His answer is correct : 2n−1 is odd, so (2n−1)!=(2n−1)(2n−3)......5.3.1 = a then he uses a = ((a×b)/b) to acheive the prof.
$$\boldsymbol{{Check}}\:\:\boldsymbol{{once}}\:\:\boldsymbol{{more}}\:\:\boldsymbol{{brother}}; \\ $$$$\boldsymbol{{His}}\:\:\boldsymbol{{answer}}\:\:\boldsymbol{{is}}\:\:\boldsymbol{{correct}}\::\:\:\mathrm{2}\boldsymbol{{n}}−\mathrm{1}\:\:\boldsymbol{{is}}\:\:\boldsymbol{{odd}},\:\:\boldsymbol{{so}}\:\:\left(\mathrm{2}\boldsymbol{{n}}−\mathrm{1}\right)!=\left(\mathrm{2}\boldsymbol{{n}}−\mathrm{1}\right)\left(\mathrm{2}\boldsymbol{{n}}−\mathrm{3}\right)……\mathrm{5}.\mathrm{3}.\mathrm{1}\:=\:\boldsymbol{{a}} \\ $$$$\boldsymbol{{then}}\:\:\boldsymbol{{he}}\:\:\boldsymbol{{uses}}\:\:\boldsymbol{{a}}\:=\:\frac{\boldsymbol{{a}}×\boldsymbol{{b}}}{\boldsymbol{{b}}}\:\:\boldsymbol{{to}}\:\:\boldsymbol{{acheive}}\:\:\boldsymbol{{the}}\:\:\boldsymbol{{prof}}. \\ $$
Commented by Tanmay chaudhury last updated on 05/Nov/19
1!=1 2!=2 3!=6 4!=24 (1!+2!+3!+4!)=33 factorial 5! and and other terms upto 99! contains=2×5=so zero... last digit of 5!=0 last digit of 6!=0 so last digit of (1!+2!+...+99!)=3 pls check...
$$\mathrm{1}!=\mathrm{1} \\ $$$$\mathrm{2}!=\mathrm{2} \\ $$$$\mathrm{3}!=\mathrm{6} \\ $$$$\mathrm{4}!=\mathrm{24}\:\:\left(\mathrm{1}!+\mathrm{2}!+\mathrm{3}!+\mathrm{4}!\right)=\mathrm{33} \\ $$$${factorial}\:\mathrm{5}!\:{and}\:{and}\:{other}\:{terms}\:{upto}\:\mathrm{99}! \\ $$$${contains}=\mathrm{2}×\mathrm{5}={so}\:{zero}… \\ $$$${last}\:{digit}\:{of}\:\mathrm{5}!=\mathrm{0}\: \\ $$$${last}\:{digit}\:{of}\:\mathrm{6}!=\mathrm{0} \\ $$$$ \\ $$$${so}\:{last}\:{digit}\:{of}\:\left(\mathrm{1}!+\mathrm{2}!+…+\mathrm{99}!\right)=\mathrm{3} \\ $$$${pls}\:{check}… \\ $$$$ \\ $$
Commented by Raxreedoroid last updated on 05/Nov/19
ok thanks, I think my misunderstanding was because I prove it in another way
$$\mathrm{ok}\:\mathrm{thanks},\:\mathrm{I}\:\mathrm{think}\:\mathrm{my}\:\mathrm{misunderstanding}\:\mathrm{was}\:\mathrm{because}\:\mathrm{I}\:\mathrm{prove}\:\mathrm{it}\:\mathrm{in}\:\mathrm{another}\:\mathrm{way} \\ $$

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