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Question Number 115095 by aurpeyz last updated on 23/Sep/20

Answered by Dwaipayan Shikari last updated on 23/Sep/20

$$Termnth=6n-1$$$$\underset{n=1}{\sum ^n}6n-1=3n(n+1)-n$$$$3n(n+1)-n>1000$$$$3n^2+2n-1000>0$$$$n^2+\frac{2n}{3}-\frac{1000}{3}>0$$$${(n+\frac{1}{3})}^2-\frac{1}{9}-\frac{1000}{3}>0$$$${(n+\frac{1}{3})}^2>\frac{3001}{9}$$$$n>-\frac{1}{3}+\sqrt{\frac{3001}{9}}$$$$n>17.92$$$$Sonhastobe18$$

Commented by aurpeyz last updated on 23/Sep/20

$$plswhatis3n(n+1)-n?$$

Commented by Dwaipayan Shikari last updated on 23/Sep/20

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{n}}}6\mathrm{n}-1=6.\frac{\mathrm{n}(\mathrm{n}+1)}{2}-\mathrm{n}$$

Answered by Aziztisffola last updated on 23/Sep/20

$${\mathrm{u}}_0=5\mathrm{\&}{\mathrm{u}}_{\mathrm{n}+1}={\mathrm{u}}_{\mathrm{n}}+6$$$$\Rightarrow {\mathrm{u}}_{\mathrm{n}}=6\mathrm{n}+5$$$$\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{u}}_{\mathrm{k}}=\frac{{\mathrm{u}}_0+{\mathrm{u}}_{\mathrm{n}}}{2}(\mathrm{n}+1)>1000$$$$(3\mathrm{n}+5)(\mathrm{n}+1)>1000$$$${3\mathrm{n}}^2+8\mathrm{n}-995>0$$$$\mathrm{solve}\mathrm{for}\mathrm{n}\in \mathbb{N}$$$$\Rightarrow \mathrm{n}>\frac{\sqrt{3001}}{2}-\frac{4}{3}=17.....$$$$\Rightarrow \mathrm{n}=18$$

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