In-each-week-the-growth-of-a-plant-is-two-thirds-the-growth-of-the-previous-week-The-plant-grows-12-cm-in-the-first-week-a-Calculate-the-growth-of-the-plant-in-b-the-limiting-height-of-the-pant

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Question Number 97483 by Rio Michael last updated on 08/Jun/20

$$\mathrm{In}\mathrm{each}\mathrm{week}\mathrm{the}\mathrm{growth}\mathrm{of}\mathrm{a}\mathrm{plant}\mathrm{is}\mathrm{two}-\mathrm{thirds}$$$$\mathrm{the}\mathrm{growth}\mathrm{of}\mathrm{the}\mathrm{previous}\mathrm{week}.$$$$\mathrm{The}\mathrm{plant}\mathrm{grows}12\mathrm{cm}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{week}.$$$$\left(\mathrm{a}\right)\mathrm{Calculate}\mathrm{the}\mathrm{growth}\mathrm{of}\mathrm{the}\mathrm{plant}\mathrm{in}$$$$\left(\mathrm{b}\right)\mathrm{the}\mathrm{limiting}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{pant}$$

Commented by bobhans last updated on 08/Jun/20

$$\left(\mathrm{a}\right)\mathrm{calculate}\mathrm{the}\mathrm{growth}\mathrm{of}\mathrm{the}\mathrm{plant}\mathrm{in}\dots ???$$$$\mathrm{not}\mathrm{clear}$$

Commented by Rio Michael last updated on 08/Jun/20

$$\mathrm{in}\mathrm{the}8\mathrm{th}\mathrm{week}\mathrm{sorry}$$

Commented by bobhans last updated on 08/Jun/20

$$\mathrm{let}{\mathrm{h}}_1=12\mathrm{cm},{\mathrm{h}}_2=12(1+\frac{2}{3})=12\left(\frac{5}{3}\right)=20\mathrm{cm}$$$${\mathrm{h}}_3=20(1+\frac{2}{3})=20\times \left(\frac{5}{3}\right)=\frac{100}{3}$$$$\mathrm{so}{\mathrm{h}}_8=12{(1+\frac{2}{3})}^8\mathrm{cm}$$

Commented by mr W last updated on 08/Jun/20

$$impossible!after8weeks{it}^{\prime }s7.14m!$$$$andafter12weeks,i.e.3monthes$$$${it}^{\prime }s55.12m!after5monthes{it}^{\prime }s$$$$3282m!$$

Answered by mr W last updated on 09/Jun/20

$$week1:growth12,height12$$$$week2:growth12\times \frac{2}{3},height12(1+\frac{2}{3})$$$$week3:growth12\times {\left(\frac{2}{3}\right)}^2,height12[1+\frac{2}{3}+{\left(\frac{2}{3}\right)}^2]$$$$\dots$$$$weekn:growth12\times {\left(\frac{2}{3}\right)}^{n-1},height12[1+\frac{2}{3}+{\left(\frac{2}{3}\right)}^2+..+{\left(\frac{2}{3}\right)}^{n-1}]$$$$heightafternweeks:h_n=12[1+\frac{2}{3}+{\left(\frac{2}{3}\right)}^2+..+{\left(\frac{2}{3}\right)}^{n-1}]=\frac{12[1-{\left(\frac{2}{3}\right)}^n]}{1-\frac{2}{3}}$$$$\Rightarrow h_n=36[1-{\left(\frac{2}{3}\right)}^n]$$$$h_8=36[1-{\left(\frac{2}{3}\right)}^8]=34.6cm$$$$$$$$h_{max}=\underset{n\to \mathrm{\infty }}{\lim }{h}_n=36cm$$

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