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Question Number 128369 by I want to learn more last updated on 06/Jan/21

$$\mathrm{If}{\mathrm{u}}_1+{\mathrm{u}}_2+{\mathrm{u}}_3+...+{\mathrm{u}}_{\mathrm{n}}={2\mathrm{n}}^2+\mathrm{n}\mathrm{is}\mathrm{an}\mathrm{AP}.$$$$\mathrm{Find}{\mathrm{u}}_1+{\mathrm{u}}_2+{\mathrm{u}}_3+...+{\mathrm{u}}_{2\mathrm{n}-2}+{\mathrm{u}}_{2\mathrm{n}-1}$$

Answered by mr W last updated on 06/Jan/21

$${\mathrm{u}}_1+{\mathrm{u}}_2+{\mathrm{u}}_3+...+{\mathrm{u}}_{2\mathrm{n}-2}+{\mathrm{u}}_{2\mathrm{n}-1}$$$$=2{(2n-1)}^2+(2n-1)$$$$=(2n-1)(4n-1)$$$$=8n^2-6n+1$$

Commented by I want to learn more last updated on 06/Jan/21

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

Answered by mathmax by abdo last updated on 06/Jan/21

$$\mathrm{we}\mathrm{have}{\mathrm{u}}_1+{\mathrm{u}}_2+...+{\mathrm{u}}_{\mathrm{n}}={2\mathrm{n}}^2+\mathrm{n}\Rightarrow$$$$\{\begin{array}{l}{\mathrm{u}}_1+{\mathrm{u}}_2+...+{\mathrm{u}}_{\mathrm{n}}={2\mathrm{n}}^2+\mathrm{n}\\ {\mathrm{u}}_1+{\mathrm{u}}_2+..{\mathrm{u}}_{\mathrm{n}-1}=2{(\mathrm{n}-1)}^2+\mathrm{n}-1\Rightarrow \end{array}$$$${\mathrm{u}}_{\mathrm{n}}={2\mathrm{n}}^2+\mathrm{n}-2{(\mathrm{n}-1)}^2-\mathrm{n}+1={2\mathrm{n}}^2-2({\mathrm{n}}^2-2\mathrm{n}+1)+1$$$$=4\mathrm{n}-2+1=4\mathrm{n}-1\mathrm{so}$$$${\mathrm{u}}_1+{\mathrm{u}}_2+....+{\mathrm{u}}_{2\mathrm{n}-1}=\frac{2\mathrm{n}-1-1+1}{2}({\mathrm{u}}_1+{\mathrm{u}}_{2\mathrm{n}-1})$$$$=\frac{2\mathrm{n}-1}{2}(3+4(2\mathrm{n}-1)-1)=\frac{2\mathrm{n}-1}{2}(3+8\mathrm{n}-4-1)$$$$=\frac{(2\mathrm{n}-1)(8\mathrm{n}-2)}{2}=(2\mathrm{n}-1)(4\mathrm{n}-1)$$$$={8\mathrm{n}}^2-2\mathrm{n}-4\mathrm{n}+1={8\mathrm{n}}^2-6\mathrm{n}+1$$

Commented by I want to learn more last updated on 07/Jan/21

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

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