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Question Number 98215 by I want to learn more last updated on 12/Jun/20

$$\mathbf{Find}\mathbf{the}\mathbf{nth}\mathbf{term}\mathbf{of}\mathbf{the}\mathbf{sequence}\left\{{\mathbf{a}}_{\mathbf{n}}\right\}\mathbf{such}\mathbf{that}$$$$\frac{{\mathbf{a}}_1+{\mathbf{a}}_2+...+{\mathbf{a}}_{\mathbf{n}}}{\mathbf{n}}=\mathbf{n}+\frac{1}{\mathbf{n}}(\mathbf{n}=1,2,3,...)$$

Commented by Don08q last updated on 12/Jun/20

$$$$$$a_1+a_2+...+a_n=n(n+\frac{1}{n})$$$$S_n=n(n+\frac{1}{n})$$$$S_n=n^2+1$$$$\Rightarrow S_{n-1}={(n-1)}^2+1$$$$\mathrm{The}nth\mathrm{term},a_n=S_n-S_{n-1}$$$$\mathrm{So},a_n=(n^2+1)-(n^2-2n+2)$$$$\therefore a_n=2n-1$$$$$$

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

$$2n-1$$

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

$$\frac{S_n}{n}=n+\frac{1}{n}=\frac{n^2+1}{n}$$$$S_n=n^2+1$$$$S_{n-1}={(n-1)}^2+1$$$$a_n=S_n-S_{n-1}=n^2+1-{(n-1)}^2-1=2n-1$$

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

$$\frac{{\mathrm{a}}_1+{\mathrm{a}}_2+....+{\mathrm{a}}_{\mathrm{n}}}{\mathrm{n}}=\mathrm{n}+\frac{1}{\mathrm{n}}\Rightarrow \sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{a}}_{\mathrm{k}}={\mathrm{n}}^2+1\Rightarrow$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}-1}{\mathrm{a}}_{\mathrm{k}}={(\mathrm{n}-1)}^2+1\Rightarrow \sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{a}}_{\mathrm{k}}-\sum _{\mathrm{k}=1}^{\mathrm{n}-1}{\mathrm{a}}_{\mathrm{k}}={\mathrm{n}}^2-{(\mathrm{n}-1)}^2\Rightarrow$$$${\mathrm{a}}_{\mathrm{n}}={\mathrm{n}}^2-({\mathrm{n}}^2-2\mathrm{n}+1)=2\mathrm{n}-1$$

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

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

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