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Question-7630

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Question Number 7630 by sandy_suhendra last updated on 06/Sep/16
Commented by Chantria Math last updated on 06/Sep/16
There are some false in this problem,
$${There}\:{are}\:{some}\:{false}\:{in}\:{this} \\ $$$${problem},\: \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by Rasheed Soomro last updated on 07/Sep/16
What false?
$${What}\:{false}? \\ $$
Commented by Rasheed Soomro last updated on 07/Sep/16
I don′t see ′upon-line′ in above question(image), it was present when at first I start the app. Now it disappears. What′s the problem?
$${I}\:{don}'{t}\:{see}\:'{upon}-{line}'\:{in}\:{above}\:{question}\left({image}\right), \\ $$$${it}\:{was}\:{present}\:{when}\:{at}\:{first}\:{I}\:{start}\:{the}\:{app}. \\ $$$${Now}\:{it}\:{disappears}.\:{What}'{s}\:{the}\:{problem}? \\ $$
Commented by sandy_suhendra last updated on 07/Sep/16
may be because in the first time, I made a mistake in type the question and then I corrected it
$${may}\:{be}\:{because}\:{in}\:{the}\:{first}\:{time},\:{I}\:{made}\:{a}\:{mistake}\:{in}\:{type}\:{the}\:{question}\:{and}\:{then}\:{I}\:{corrected}\:{it} \\ $$
Commented by Rasheed Soomro last updated on 07/Sep/16
It′s the position even now after your correction! When I start the app I can see the upon-line but if I minimize and then maximize the upon-line disappears!!
$${It}'{s}\:{the}\:{position}\:{even}\:{now}\:{after}\:{your}\:{correction}! \\ $$$${When}\:{I}\:{start}\:{the}\:{app}\:{I}\:{can}\:{see}\:{the}\:{upon}-{line} \\ $$$${but}\:{if}\:{I}\:{minimize}\:{and}\:{then}\:{maximize}\:{the}\:{upon}-{line} \\ $$$${disappears}!! \\ $$
Answered by Yozzia last updated on 06/Sep/16
The given expression is equal to (1/(24)). −−−−−−−−−−−−−−−−−−−−−−−− Let u=((Π_(i=1) ^8 (3^2^i +1))/(27^(171) −3))=((Π_(i=1) ^8 (3^2^i +1))/(3^(513) −3))=((Π_(i=1) ^8 (3^2^i +1))/(3(3^(512) −1))) Now, 3^(512) −1=(3^(256) −1)(3^(256) +1) =(3^(128) −1)(3^(128) +1)(3^(256) +1) =(3^(64) −1)(3^(64) +1)(3^(128) +1)(3^(256) +1) =(3^(32) −1)(3^(32) +1)(3^(64) +1)(3^(128) +1)(3^(256) +1) =(3^(16) −1)Π_(i=4) ^8 (3^2^i +1) =(3^8 −1)Π_(i=3) ^8 (3^2^i +1) =(3^4 −1)Π_(i=2) ^8 (3^2^i +1) =(3^2 −1)Π_(i=1) ^8 (3^2^i +1) 3^(512) −1=8×Π_(i=1) ^8 (3^2^i +1) ∴ u=((Π_(i=1) ^8 (3^2^i +1))/(3×8×Π_(i=1) ^8 (3^2^i +1)))=(1/(24))
$${The}\:{given}\:{expression}\:{is}\:{equal}\:{to}\:\frac{\mathrm{1}}{\mathrm{24}}. \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${Let}\:{u}=\frac{\underset{{i}=\mathrm{1}} {\overset{\mathrm{8}} {\prod}}\left(\mathrm{3}^{\mathrm{2}^{{i}} } +\mathrm{1}\right)}{\mathrm{27}^{\mathrm{171}} −\mathrm{3}}=\frac{\underset{{i}=\mathrm{1}} {\overset{\mathrm{8}} {\prod}}\left(\mathrm{3}^{\mathrm{2}^{{i}} } +\mathrm{1}\right)}{\mathrm{3}^{\mathrm{513}} −\mathrm{3}}=\frac{\underset{{i}=\mathrm{1}} {\overset{\mathrm{8}} {\prod}}\left(\mathrm{3}^{\mathrm{2}^{{i}} } +\mathrm{1}\right)}{\mathrm{3}\left(\mathrm{3}^{\mathrm{512}} −\mathrm{1}\right)} \\ $$$${Now},\:\mathrm{3}^{\mathrm{512}} −\mathrm{1}=\left(\mathrm{3}^{\mathrm{256}} −\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{256}} +\mathrm{1}\right) \\ $$$$=\left(\mathrm{3}^{\mathrm{128}} −\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{128}} +\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{256}} +\mathrm{1}\right) \\ $$$$=\left(\mathrm{3}^{\mathrm{64}} −\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{64}} +\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{128}} +\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{256}} +\mathrm{1}\right) \\ $$$$=\left(\mathrm{3}^{\mathrm{32}} −\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{32}} +\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{64}} +\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{128}} +\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{256}} +\mathrm{1}\right) \\ $$$$=\left(\mathrm{3}^{\mathrm{16}} −\mathrm{1}\right)\underset{{i}=\mathrm{4}} {\overset{\mathrm{8}} {\prod}}\left(\mathrm{3}^{\mathrm{2}^{{i}} } +\mathrm{1}\right) \\ $$$$=\left(\mathrm{3}^{\mathrm{8}} −\mathrm{1}\right)\underset{{i}=\mathrm{3}} {\overset{\mathrm{8}} {\prod}}\left(\mathrm{3}^{\mathrm{2}^{{i}} } +\mathrm{1}\right) \\ $$$$=\left(\mathrm{3}^{\mathrm{4}} −\mathrm{1}\right)\underset{{i}=\mathrm{2}} {\overset{\mathrm{8}} {\prod}}\left(\mathrm{3}^{\mathrm{2}^{{i}} } +\mathrm{1}\right) \\ $$$$=\left(\mathrm{3}^{\mathrm{2}} −\mathrm{1}\right)\underset{{i}=\mathrm{1}} {\overset{\mathrm{8}} {\prod}}\left(\mathrm{3}^{\mathrm{2}^{{i}} } +\mathrm{1}\right) \\ $$$$\mathrm{3}^{\mathrm{512}} −\mathrm{1}=\mathrm{8}×\underset{{i}=\mathrm{1}} {\overset{\mathrm{8}} {\prod}}\left(\mathrm{3}^{\mathrm{2}^{{i}} } +\mathrm{1}\right) \\ $$$$\therefore\:{u}=\frac{\underset{{i}=\mathrm{1}} {\overset{\mathrm{8}} {\prod}}\left(\mathrm{3}^{\mathrm{2}^{{i}} } +\mathrm{1}\right)}{\mathrm{3}×\mathrm{8}×\underset{{i}=\mathrm{1}} {\overset{\mathrm{8}} {\prod}}\left(\mathrm{3}^{\mathrm{2}^{{i}} } +\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{24}} \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 07/Sep/16
NICE !
$$\mathcal{NICE}\:! \\ $$
Commented by sandy_suhendra last updated on 08/Sep/16
great, thank′s a lot
$${great},\:{thank}'{s}\:{a}\:{lot} \\ $$

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