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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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