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Question Number 98205 by I want to learn more last updated on 12/Jun/20
Find the nth term of the sequence {a_n } such that a_1 = 1, a_(n + 1) = (1/2)a_n + ((n^2 − 2n − 1)/(n^2 (n + 1)^2 )) (n = 1, 2, 3, ...)
$$\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{nth}}\:\:\boldsymbol{\mathrm{term}}\:\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sequence}}\:\:\left\{\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \right\}\:\:\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}} \\ $$$$\boldsymbol{\mathrm{a}}_{\mathrm{1}} \:\:=\:\:\mathrm{1},\:\:\:\:\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}\:\:+\:\:\mathrm{1}} \:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \:\:+\:\:\frac{\boldsymbol{\mathrm{n}}^{\mathrm{2}} \:−\:\mathrm{2}\boldsymbol{\mathrm{n}}\:\:−\:\:\mathrm{1}}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{n}}\:\:+\:\:\mathrm{1}\right)^{\mathrm{2}} }\:\:\:\:\left(\boldsymbol{\mathrm{n}}\:\:=\:\:\mathrm{1},\:\:\mathrm{2},\:\:\mathrm{3},\:\:…\right) \\ $$
Answered by mr W last updated on 12/Jun/20
((n^2 −2n−1)/(n^2 (n+1)^2 )) =((2n^2 −(n+1)^2 )/(n^2 (n+1)^2 )) =(2/((n+1)^2 ))−(1/n^2 ) ⇒a_(n+1) =(1/2)a_n +(2/((n+1)^2 ))−(1/n^2 ) ⇒a_(n+1) −(2/((n+1)^2 ))=(1/2)(a_n −(2/n^2 )) let b_n =a_n −(2/n^2 ) ⇒b_(n+1) =(1/2)b_n ⇒ G.P. ⇒b_n =b_1 ((1/2))^(n−1) =(a_1 −(2/1^1 ))(1/2^(n−1) )=−(1/2^(n−1) ) ⇒a_n =−(1/2^(n−1) )+(2/n^2 )=2((1/n^2 )−(1/2^n ))
$$\frac{{n}^{\mathrm{2}} −\mathrm{2}{n}−\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{n}^{\mathrm{2}} −\left({n}+\mathrm{1}\right)^{\mathrm{2}} }{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\Rightarrow{a}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}{a}_{{n}} +\frac{\mathrm{2}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\Rightarrow{a}_{{n}+\mathrm{1}} −\frac{\mathrm{2}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\left({a}_{{n}} −\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\right) \\ $$$${let}\:{b}_{{n}} ={a}_{{n}} −\frac{\mathrm{2}}{{n}^{\mathrm{2}} } \\ $$$$\Rightarrow{b}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}{b}_{{n}} \:\Rightarrow\:{G}.{P}. \\ $$$$\Rightarrow{b}_{{n}} ={b}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} =\left({a}_{\mathrm{1}} −\frac{\mathrm{2}}{\mathrm{1}^{\mathrm{1}} }\right)\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }=−\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$$$\Rightarrow{a}_{{n}} =−\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }+\frac{\mathrm{2}}{{n}^{\mathrm{2}} }=\mathrm{2}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right) \\ $$
Commented by I want to learn more last updated on 12/Jun/20
Thanks sir
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$
Commented by I want to learn more last updated on 12/Jun/20
Ok, each time, i will think how to transform it to linear.
$$\mathrm{Ok},\:\mathrm{each}\:\mathrm{time},\:\mathrm{i}\:\mathrm{will}\:\mathrm{think}\:\mathrm{how}\:\mathrm{to}\:\mathrm{transform}\:\mathrm{it}\:\mathrm{to}\:\mathrm{linear}. \\ $$
Commented by Rio Michael last updated on 12/Jun/20
sir please any hint on how to approach Q 98208??
$$\:\mathrm{sir}\:\mathrm{please}\:\mathrm{any}\:\mathrm{hint}\:\mathrm{on}\:\mathrm{how}\:\mathrm{to}\:\mathrm{approach}\:{Q}\:\mathrm{98208}?? \\ $$$$ \\ $$

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