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2-3-2-18-2-27-2-324-




Question Number 106503 by Dwaipayan Shikari last updated on 05/Aug/20
(2/3)+(2/(18))+(2/(27))+(2/(324))+....
$$\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{18}}+\frac{\mathrm{2}}{\mathrm{27}}+\frac{\mathrm{2}}{\mathrm{324}}+…. \\ $$
Commented by Dwaipayan Shikari last updated on 05/Aug/20
2((1/(2+1))+(1/(2(2+1)^2 ))+(1/(3.(2+1)^3 ))+....)  =2((1/x)+(1/(2x^2 ))+(1/(3x^3 ))+....)  =−2log(1−(1/x))=−2log((2/3))  =log((9/4))   x=3  =0.81093..
$$\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}.\left(\mathrm{2}+\mathrm{1}\right)^{\mathrm{3}} }+….\right) \\ $$$$=\mathrm{2}\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }+….\right) \\ $$$$=−\mathrm{2}{log}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)=−\mathrm{2}{log}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\:\:={log}\left(\frac{\mathrm{9}}{\mathrm{4}}\right)\:\:\:{x}=\mathrm{3} \\ $$$$=\mathrm{0}.\mathrm{81093}.. \\ $$
Commented by Ar Brandon last updated on 05/Aug/20
I think you changed the question.�� I struggled hard to establish a general term.
Commented by mr W last updated on 05/Aug/20
2+(2/(11))+(2/(25))+(2/(69))+...=?
$$\mathrm{2}+\frac{\mathrm{2}}{\mathrm{11}}+\frac{\mathrm{2}}{\mathrm{25}}+\frac{\mathrm{2}}{\mathrm{69}}+…=? \\ $$
Commented by Dwaipayan Shikari last updated on 05/Aug/20
Oh I had changed the question. Sorry for the inconvenience ��
Commented by Ar Brandon last updated on 05/Aug/20
��Mr W
Commented by Dwaipayan Shikari last updated on 05/Aug/20
Great question
$${Great}\:{question} \\ $$
Commented by mr W last updated on 05/Aug/20
i  was just wondering how you can find  the general term through just four  given terms! for me it is almost  impossible to see the general term,  for example my example above.
$${i}\:\:{was}\:{just}\:{wondering}\:{how}\:{you}\:{can}\:{find} \\ $$$${the}\:{general}\:{term}\:{through}\:{just}\:{four} \\ $$$${given}\:{terms}!\:{for}\:{me}\:{it}\:{is}\:{almost} \\ $$$${impossible}\:{to}\:{see}\:{the}\:{general}\:{term}, \\ $$$${for}\:{example}\:{my}\:{example}\:{above}. \\ $$
Commented by prakash jain last updated on 05/Aug/20
Suppose 1..k terms of a sequence  are given and formula  g(n) satisifies the given terms.  then f(n)=h(n)Π_(j=1) ^k (n−k)+g(n)  will also satify the given sequence.  where h(n) is any arbitary function  on n.  So there always exists infinite number  of formula for a given finite terms.
$$\mathrm{Suppose}\:\mathrm{1}..{k}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{a}\:\mathrm{sequence} \\ $$$$\mathrm{are}\:\mathrm{given}\:\mathrm{and}\:\mathrm{formula} \\ $$$${g}\left({n}\right)\:\mathrm{satisifies}\:\mathrm{the}\:\mathrm{given}\:\mathrm{terms}. \\ $$$$\mathrm{then}\:{f}\left({n}\right)={h}\left({n}\right)\underset{{j}=\mathrm{1}} {\overset{{k}} {\prod}}\left({n}−{k}\right)+{g}\left({n}\right) \\ $$$$\mathrm{will}\:\mathrm{also}\:\mathrm{satify}\:\mathrm{the}\:\mathrm{given}\:\mathrm{sequence}. \\ $$$$\mathrm{where}\:{h}\left({n}\right)\:\mathrm{is}\:\mathrm{any}\:\mathrm{arbitary}\:\mathrm{function} \\ $$$$\mathrm{on}\:{n}. \\ $$$$\mathrm{So}\:\mathrm{there}\:\mathrm{always}\:\mathrm{exists}\:\mathrm{infinite}\:\mathrm{number} \\ $$$$\mathrm{of}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{a}\:\mathrm{given}\:\mathrm{finite}\:\mathrm{terms}. \\ $$
Commented by Ar Brandon last updated on 06/Aug/20
��
Commented by 1549442205PVT last updated on 06/Aug/20
We can  find the general term in form  (2/a_n )with a_n =f(x)=ax^3 +bx^2 +cx+d with condition  that f(1)=1,f(2)=11,f(3)=25,f(4)=69  which lead to solve the system of four  unknows a,b,c,d and get  a=((13)/3),b=−24,c=((155)/3),d=−31and so  a_n =((13)/3)n^3 −24n^2 +155n−31which from  this we get the next terms of the given  sequence:(2/(169)),(2/(351)),(2/(641)),(2/(1065)),...and   Of course,here need must be suppose  that any  formular  was found out only  need to correct for four given terms  because otherwise  we can find a    polynomial of degree 5 which also get   four first values of given sequence  Which this completely depend on the   viewer pointof person give the question
$$\mathrm{We}\:\mathrm{can}\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{general}\:\mathrm{term}\:\mathrm{in}\:\mathrm{form} \\ $$$$\frac{\mathrm{2}}{\mathrm{a}_{\mathrm{n}} }\mathrm{with}\:\mathrm{a}_{\mathrm{n}} =\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ax}^{\mathrm{3}} +\mathrm{bx}^{\mathrm{2}} +\mathrm{cx}+\mathrm{d}\:\mathrm{with}\:\mathrm{condition} \\ $$$$\mathrm{that}\:\mathrm{f}\left(\mathrm{1}\right)=\mathrm{1},\mathrm{f}\left(\mathrm{2}\right)=\mathrm{11},\mathrm{f}\left(\mathrm{3}\right)=\mathrm{25},\mathrm{f}\left(\mathrm{4}\right)=\mathrm{69} \\ $$$$\mathrm{which}\:\mathrm{lead}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{four} \\ $$$$\mathrm{unknows}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\mathrm{and}\:\mathrm{get} \\ $$$$\mathrm{a}=\frac{\mathrm{13}}{\mathrm{3}},\mathrm{b}=−\mathrm{24},\mathrm{c}=\frac{\mathrm{155}}{\mathrm{3}},\mathrm{d}=−\mathrm{31and}\:\mathrm{so} \\ $$$$\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{13}}{\mathrm{3}}\mathrm{n}^{\mathrm{3}} −\mathrm{24n}^{\mathrm{2}} +\mathrm{155n}−\mathrm{31which}\:\mathrm{from} \\ $$$$\mathrm{this}\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{next}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{sequence}:\frac{\mathrm{2}}{\mathrm{169}},\frac{\mathrm{2}}{\mathrm{351}},\frac{\mathrm{2}}{\mathrm{641}},\frac{\mathrm{2}}{\mathrm{1065}},…\mathrm{and}\: \\ $$$$\mathrm{Of}\:\mathrm{course},\mathrm{here}\:\mathrm{need}\:\mathrm{must}\:\mathrm{be}\:\mathrm{suppose} \\ $$$$\mathrm{that}\:\mathrm{any}\:\:\mathrm{formular}\:\:\mathrm{was}\:\mathrm{found}\:\mathrm{out}\:\mathrm{only} \\ $$$$\mathrm{need}\:\mathrm{to}\:\mathrm{correct}\:\mathrm{for}\:\mathrm{four}\:\mathrm{given}\:\mathrm{terms} \\ $$$$\mathrm{because}\:\mathrm{otherwise}\:\:\mathrm{we}\:\mathrm{can}\:\mathrm{find}\:\mathrm{a}\:\: \\ $$$$\mathrm{polynomial}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{5}\:\mathrm{which}\:\mathrm{also}\:\mathrm{get}\: \\ $$$$\mathrm{four}\:\mathrm{first}\:\mathrm{values}\:\mathrm{of}\:\mathrm{given}\:\mathrm{sequence} \\ $$$$\mathrm{Which}\:\mathrm{this}\:\mathrm{completely}\:\mathrm{depend}\:\mathrm{on}\:\mathrm{the}\: \\ $$$$\mathrm{viewer}\:\mathrm{pointof}\:\mathrm{person}\:\mathrm{give}\:\mathrm{the}\:\mathrm{question} \\ $$$$ \\ $$
Commented by Ar Brandon last updated on 06/Aug/20
Wow, thanks for the knowledge Sir.

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