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Question Number 5744 by Rasheed Soomro last updated on 26/May/16

$$If0<r<1,isSconvergentinthefollowing?$$ $$S=a^2+ar(a+r)+{ar}^2(a+2r)+.....$$ $$DetermineSif{it}^{\prime }sconvergent.$$

Commented byFilupSmith last updated on 26/May/16

$$\mathrm{Yes}.\mathrm{if}r<1,r=\frac{1}{n}$$ $$\frac{1}{n}>\frac{1}{n^2}>...$$ $$\mathrm{each}\mathrm{terms}\mathrm{becomes}\mathrm{smaller}$$ $$----------$$ $$S=a^2+ar(a+r)+{ar}^2(a+2r)+...$$ $$S=a^2+a\underset{i=1}{\sum ^m}{r}^n(a+nr)$$ $$T=\underset{i=1}{\sum ^m}{r}^n(a+nr)$$ $$T=r(a+r)+r^2(a+2r)+r^3(a+3r)+...$$ $$T=ar+r^2+{ar}^2+2r^3+{ar}^3+3r^4+...$$ $$T=(ar+{ar}^2+...+{ar}^m)+(r^2+2r^3+3r^4+...+(m-1)r^m)$$ $$T=a(r+r^2+...+r^m)+(r^2+2r^3+3r^4+...+(m-1)r^m)$$ $$T=a\underset{i=1}{\sum ^m}{r}^i+r\underset{i=1}{\sum ^{m-1}}{ir}^i$$ $$a\underset{i=1}{\sum ^m}{r}^i=a\left(\frac{r(1-r^m)}{1-r}\right)r<1$$ $$$$ $$S=a^2+a\underset{i=1}{\sum ^m}{r}^n(a+nr)$$ $$S=a^2+T$$ $$\therefore S=a^2+\frac{ar(1-r^m)}{1-r}+r\underset{i=1}{\sum ^m}{ir}^i,r<1$$ $$\left(\mathrm{continue}\right)$$

Commented byFilupSmith last updated on 26/May/16

$$\mathrm{I}\mathrm{am}\mathrm{sure}\mathrm{my}\mathrm{working}\mathrm{is}\mathrm{correct},\mathrm{but}$$ $$\mathrm{I}\mathrm{am}\mathrm{unsure}\mathrm{how}\mathrm{to}\mathrm{continue}$$

Commented byRasheed Soomro last updated on 26/May/16

$$\mathrm{I}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{consider}\mathrm{about}\mathrm{the}\mathrm{solution}.\mathrm{However}\mathrm{I}\mathrm{want}\mathrm{to}\mathrm{share}$$ $$\text{You can't use 'macro parameter character \#' in math mode}$$ $$$$ $$\mathrm{I}\mathrm{have}\mathrm{created}\mathrm{these}\mathrm{questions}\mathrm{as}\mathrm{follows}:$$ $$\mathrm{Consider}\mathrm{an}\mathrm{A}.\mathrm{P}.\mathrm{and}\mathrm{a}\mathrm{G}.\mathrm{P}.\mathrm{having}\mathrm{first}\mathrm{term}\mathrm{same}$$ $$\left(\mathrm{say}a\right)\mathrm{and}\mathrm{equal}\mathrm{common}\mathrm{difference}\mathrm{and}$$ $$\mathrm{common}\mathrm{ratio}\left(\mathrm{say}r\right).$$ $$a,a+r,a+2r....\mathrm{\&}a,ar,{ar}^2....$$ $$\mathrm{Now}\mathrm{consider}\mathrm{a}\mathrm{rectangle}\mathrm{whose}\mathrm{dimentions}$$ $$\mathrm{are}\mathrm{corresponding}\mathrm{terms}\mathrm{of}\mathrm{above}\mathrm{sequence}.$$ $$\mathrm{All}\mathrm{such}\mathrm{rectangles}\mathrm{form}\mathrm{a}\mathrm{sequence}...\mathrm{Sum}$$ $$\mathrm{of}\mathrm{all}\mathrm{rectangles}....$$ $$$$

Commented byFilupSmith last updated on 26/May/16

$$\mathrm{I}\mathrm{see}.\mathrm{I}\mathrm{love}\mathrm{doing}\mathrm{these}\mathrm{problems},\mathrm{also}.$$ $$\mathrm{Especially}\mathrm{fractal}-\mathrm{like}\mathrm{shapes}.$$

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