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Question Number 98215 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 + a_2 + ... + a_n )/n) = n + (1/n) (n = 1, 2, 3, ...)

$$\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{nth}}\:\:\boldsymbol{\mathrm{term}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sequence}}\:\:\left\{\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \right\}\:\:\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}} \\ $$$$\:\:\:\:\frac{\boldsymbol{\mathrm{a}}_{\mathrm{1}} \:+\:\:\boldsymbol{\mathrm{a}}_{\mathrm{2}} \:+\:\:...\:\:+\:\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{n}}}\:\:\:=\:\:\boldsymbol{\mathrm{n}}\:\:+\:\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\:\:\left(\boldsymbol{\mathrm{n}}\:\:=\:\:\mathrm{1},\:\:\mathrm{2},\:\:\mathrm{3},\:\:...\right) \\ $$

Commented by Don08q last updated on 12/Jun/20

a_1 + a_2 + ... + a_n = n(n + (1/n)) S_n = n(n + (1/n)) S_n = n^2 + 1 ⇒ S_(n−1) = (n − 1)^2 + 1 The nth term, a_n = S_n − S_(n−1) So, a_n = (n^2 + 1) − (n^2 − 2n + 2) ∴ a_n = 2n − 1

$$ \\ $$$$\:\:{a}_{\mathrm{1}} \:+\:{a}_{\mathrm{2}} \:+\:...\:+\:{a}_{{n}} \:=\:{n}\left({n}\:+\:\frac{\mathrm{1}}{{n}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{S}_{{n}} \:=\:{n}\left({n}\:+\:\frac{\mathrm{1}}{{n}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{S}_{{n}} \:=\:{n}^{\mathrm{2}} \:+\:\mathrm{1} \\ $$$$\:\Rightarrow\:\:{S}_{{n}−\mathrm{1}} \:=\:\left({n}\:−\:\mathrm{1}\right)^{\mathrm{2}} \:+\:\mathrm{1} \\ $$$$\:\:\mathrm{The}\:{nth}\:\mathrm{term},\:{a}_{{n}} \:=\:{S}_{{n}} \:−\:{S}_{{n}−\mathrm{1}} \: \\ $$$$\:\:\mathrm{So},\:{a}_{{n}} \:=\:\left({n}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:−\:\left({n}^{\mathrm{2}} \:−\:\mathrm{2}{n}\:+\:\mathrm{2}\right) \\ $$$$\:\:\therefore\:\:\:\:{a}_{{n}} \:=\:\mathrm{2}{n}\:−\:\mathrm{1} \\ $$$$ \\ $$

Commented by I want to learn more last updated on 12/Jun/20

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Answered by aadf last updated on 12/Jun/20

2n−1

$$\mathrm{2}{n}−\mathrm{1} \\ $$

Answered by mr W last updated on 12/Jun/20

(S_n /n)=n+(1/n)=((n^2 +1)/n) S_n =n^2 +1 S_(n−1) =(n−1)^2 +1 a_n =S_n −S_(n−1) =n^2 +1−(n−1)^2 −1=2n−1

$$\frac{{S}_{{n}} }{{n}}={n}+\frac{\mathrm{1}}{{n}}=\frac{{n}^{\mathrm{2}} +\mathrm{1}}{{n}} \\ $$$${S}_{{n}} ={n}^{\mathrm{2}} +\mathrm{1} \\ $$$${S}_{{n}−\mathrm{1}} =\left({n}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$${a}_{{n}} ={S}_{{n}} −{S}_{{n}−\mathrm{1}} ={n}^{\mathrm{2}} +\mathrm{1}−\left({n}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}=\mathrm{2}{n}−\mathrm{1} \\ $$

Commented by I want to learn more last updated on 12/Jun/20

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Answered by mathmax by abdo last updated on 12/Jun/20

((a_1 +a_2 +....+a_n )/n) =n+(1/n) ⇒Σ_(k=1) ^n a_k =n^2 +1 ⇒ Σ_(k=1) ^(n−1) a_k =(n−1)^2 +1 ⇒Σ_(k=1) ^n a_k −Σ_(k=1) ^(n−1) a_k =n^2 −(n−1)^2 ⇒ a_n =n^2 −(n^2 −2n+1) =2n−1

$$\frac{\mathrm{a}_{\mathrm{1}} \:+\mathrm{a}_{\mathrm{2}} \:+....+\mathrm{a}_{\mathrm{n}} }{\mathrm{n}}\:=\mathrm{n}+\frac{\mathrm{1}}{\mathrm{n}}\:\Rightarrow\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{a}_{\mathrm{k}} =\mathrm{n}^{\mathrm{2}} +\mathrm{1}\:\Rightarrow \\ $$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \:\mathrm{a}_{\mathrm{k}} =\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\:\Rightarrow\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{a}_{\mathrm{k}} −\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \:\mathrm{a}_{\mathrm{k}} =\mathrm{n}^{\mathrm{2}} −\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{a}_{\mathrm{n}} =\mathrm{n}^{\mathrm{2}} −\left(\mathrm{n}^{\mathrm{2}} −\mathrm{2n}+\mathrm{1}\right)\:=\mathrm{2n}−\mathrm{1} \\ $$

Commented by mathmax by abdo last updated on 12/Jun/20

you are welcome

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

Commented by I want to learn more last updated on 12/Jun/20

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

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