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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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