Question Number 98215 by I want to learn more last updated on 12/Jun/20
$$\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{nth}}\:\:\boldsymbol{\mathrm{term}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sequence}}\:\:\left\{\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \right\}\:\:\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}} \\ $$$$\:\:\:\:\frac{\boldsymbol{\mathrm{a}}_{\mathrm{1}} \:+\:\:\boldsymbol{\mathrm{a}}_{\mathrm{2}} \:+\:\:…\:\:+\:\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{n}}}\:\:\:=\:\:\boldsymbol{\mathrm{n}}\:\:+\:\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\:\:\left(\boldsymbol{\mathrm{n}}\:\:=\:\:\mathrm{1},\:\:\mathrm{2},\:\:\mathrm{3},\:\:…\right) \\ $$
Commented by Don08q last updated on 12/Jun/20
$$ \\ $$$$\:\:{a}_{\mathrm{1}} \:+\:{a}_{\mathrm{2}} \:+\:…\:+\:{a}_{{n}} \:=\:{n}\left({n}\:+\:\frac{\mathrm{1}}{{n}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{S}_{{n}} \:=\:{n}\left({n}\:+\:\frac{\mathrm{1}}{{n}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{S}_{{n}} \:=\:{n}^{\mathrm{2}} \:+\:\mathrm{1} \\ $$$$\:\Rightarrow\:\:{S}_{{n}−\mathrm{1}} \:=\:\left({n}\:−\:\mathrm{1}\right)^{\mathrm{2}} \:+\:\mathrm{1} \\ $$$$\:\:\mathrm{The}\:{nth}\:\mathrm{term},\:{a}_{{n}} \:=\:{S}_{{n}} \:−\:{S}_{{n}−\mathrm{1}} \: \\ $$$$\:\:\mathrm{So},\:{a}_{{n}} \:=\:\left({n}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:−\:\left({n}^{\mathrm{2}} \:−\:\mathrm{2}{n}\:+\:\mathrm{2}\right) \\ $$$$\:\:\therefore\:\:\:\:{a}_{{n}} \:=\:\mathrm{2}{n}\:−\:\mathrm{1} \\ $$$$ \\ $$
Commented by I want to learn more last updated on 12/Jun/20
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$
Answered by aadf last updated on 12/Jun/20
$$\mathrm{2}{n}−\mathrm{1} \\ $$
Answered by mr W last updated on 12/Jun/20
$$\frac{{S}_{{n}} }{{n}}={n}+\frac{\mathrm{1}}{{n}}=\frac{{n}^{\mathrm{2}} +\mathrm{1}}{{n}} \\ $$$${S}_{{n}} ={n}^{\mathrm{2}} +\mathrm{1} \\ $$$${S}_{{n}−\mathrm{1}} =\left({n}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$${a}_{{n}} ={S}_{{n}} −{S}_{{n}−\mathrm{1}} ={n}^{\mathrm{2}} +\mathrm{1}−\left({n}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}=\mathrm{2}{n}−\mathrm{1} \\ $$
Commented by I want to learn more last updated on 12/Jun/20
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$
Answered by mathmax by abdo last updated on 12/Jun/20
$$\frac{\mathrm{a}_{\mathrm{1}} \:+\mathrm{a}_{\mathrm{2}} \:+….+\mathrm{a}_{\mathrm{n}} }{\mathrm{n}}\:=\mathrm{n}+\frac{\mathrm{1}}{\mathrm{n}}\:\Rightarrow\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{a}_{\mathrm{k}} =\mathrm{n}^{\mathrm{2}} +\mathrm{1}\:\Rightarrow \\ $$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \:\mathrm{a}_{\mathrm{k}} =\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\:\Rightarrow\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{a}_{\mathrm{k}} −\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \:\mathrm{a}_{\mathrm{k}} =\mathrm{n}^{\mathrm{2}} −\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{a}_{\mathrm{n}} =\mathrm{n}^{\mathrm{2}} −\left(\mathrm{n}^{\mathrm{2}} −\mathrm{2n}+\mathrm{1}\right)\:=\mathrm{2n}−\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 12/Jun/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$
Commented by I want to learn more last updated on 12/Jun/20
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$