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Question Number 98690 by abony1303 last updated on 15/Jun/20

Commented by abony1303 last updated on 15/Jun/20

$$\mathrm{pls}\mathrm{help}$$

Commented by PRITHWISH SEN 2 last updated on 15/Jun/20

$${\mathrm{a}}_1=3$$$${\mathrm{a}}_2=3\times 2=6$$$${\mathrm{a}}_3=6\times 2=12=3\times 2^2$$$$...................$$$${\mathrm{a}}_{\mathrm{n}}=3\times 2^{\mathrm{n}-1}$$$$\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}{\mathrm{a}}_{\mathrm{k}}=3+3\times 2+3\times 2^2+........3\times 2^{\mathrm{n}-1}=3.\frac{2^{\mathrm{n}}-1}{2-1}$$$$\mathbf{now}$$$$3(2^{\mathbf{n}}-1)>1000\Rightarrow 2^{\mathbf{n}}>\frac{1003}{3}=334\frac{1}{3}$$$$\Rightarrow \mathbf{n}=9\mathbf{please}\mathbf{check}$$

Commented by abony1303 last updated on 15/Jun/20

$$\mathrm{Thank}\mathrm{you},\mathrm{sir}!$$

Answered by Rio Michael last updated on 16/Jun/20

$$\mathrm{The}\mathrm{sequence}\left\{a_n\right\}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}{a}_{n+1}=k{a}_n$$$$\mathrm{which}\mathrm{is}\mathrm{a}\mathrm{Geometric}\mathrm{progression}\mathrm{with}\mathrm{terms}\mathrm{listed}\mathrm{below}:$$$$a_1=3,a_2=6,a_3=12,..$$$$\underset{k=1}{\sum ^n}{a}_k=S_n=\frac{a_1(r^{n-1}-1)}{r-1}\mid r\mid >1$$$$\Rightarrow \frac{3(2^{n-1}-1)}{2-1}>1000$$$$\Rightarrow 3(2^{n-1}-1)>1000$$$$\Rightarrow 2^{n-1}-1>333.33\overline{3}$$$$\Rightarrow 2^{n-1}>334.33\overline{3}$$$$\Rightarrow (n-1)\log 2>\log (334.33\overline{3})$$$$\Rightarrow (n-1)>8.385...$$$$\Rightarrow n>9.385...\Rightarrow \mathrm{smallest}\mathrm{natural}\mathrm{number}=9$$$$\mathrm{These}\mathrm{method}\mathrm{is}\mathrm{generally}\mathrm{not}\mathrm{cool}!\mathrm{it}\mathrm{is}\mathrm{possible}$$$$\mathrm{in}\mathrm{this}\mathrm{case}\mathrm{due}\mathrm{to}\mathrm{the}\mathrm{positive}\mathrm{nature}\mathrm{of}\mathrm{the}\mathrm{given}$$$$\mathrm{sequence}.$$

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