Menu Close

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-




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

Leave a Reply

Your email address will not be published. Required fields are marked *