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Question Number 8252 by 8168 last updated on 04/Oct/16

Answered by Yozzias last updated on 04/Oct/16

$$6+4+\frac{8}{3}+\frac{16}{9}+...$$$$=6+(\frac{2^2}{3^0}+\frac{2^3}{3^1}+\frac{2^4}{3^2}+...+\frac{2^{\mathrm{n}+1}}{3^{\mathrm{n}-1}}+...)$$$$=6+\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{2^{\mathrm{n}+1}}{3^{\mathrm{n}-1}}$$$$=6+\frac{2}{3^{-1}}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{\left(\frac{2}{3}\right)}^{\mathrm{n}}$$$$=6+6\times \frac{2/3}{1-\frac{2}{3}}$$$$=6+6\times \frac{2/3}{1/3}$$$$6+4+\frac{8}{3}+\frac{16}{9}+...=18$$$$\mathrm{The}\mathrm{sum}\mathrm{is}\mathrm{convergent}\mathrm{and}\mathrm{hence}$$$$\mathrm{bounded}\mathrm{and}\mathrm{monotone}.$$$$\mathrm{For}\mathrm{s}\left(\mathrm{n}\right)=6+6\underset{\mathrm{r}=1}{\sum ^{\mathrm{n}}}{\left(\frac{2}{3}\right)}^{\mathrm{r}}$$$$\Rightarrow \mathrm{s}(\mathrm{n}+1)-\mathrm{s}\left(\mathrm{n}\right)=6{\left(\frac{2}{3}\right)}^{\mathrm{n}+1}>0$$$$\Rightarrow \mathrm{s}(\mathrm{n}+1)>\mathrm{s}\left(\mathrm{n}\right)\Rightarrow \mathrm{s}\left(\mathrm{n}\right)\mathrm{is}\mathrm{an}\mathrm{increasing}$$$$\mathrm{sequence}\mathrm{of}\mathrm{partial}\mathrm{sums}.$$$$\mathrm{Since}\mathrm{s}\left(\mathrm{n}\right)\mathrm{is}\mathrm{increasing}\mathrm{and}\underset{\mathrm{n}\to \mathrm{\infty }}{\lim s}\left(\mathrm{n}\right)=18$$$$\mathrm{then}\mathrm{s}\left(\mathrm{n}\right)\mathrm{is}\mathrm{bounded}\mathrm{above}\mathrm{and}$$$$\mathrm{s}\left(\mathrm{n}\right)⩽18\mathrm{\forall }\mathrm{n}\in \mathbb{N}.$$$$$$

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