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Question Number 91796 by zainal tanjung last updated on 03/May/20

$$\mathrm{Total}\mathrm{sum}\mathrm{of}\mathrm{this}\mathrm{below}\mathrm{infinite}\mathrm{series}:$$$$1).\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...\frac{1}{\mathrm{n}(\mathrm{n}+1)}=...$$$$2).0.6+0.06+0.006+...\frac{6}{{10}^{\mathrm{n}}}=...$$

Commented by mathmax by abdo last updated on 04/May/20

$$2)S=\sum _{k=1}^n\frac{6}{{10}^k}=\frac{6}{10}\sum _{k=1}^n\frac{1}{{10}^{k-1}}=\frac{6}{10}\sum _{p=0}^{n-1}{\left(\frac{1}{10}\right)}^p$$$$=\frac{6}{10}\times \frac{1-{\left(\frac{1}{10}\right)}^n}{1-\frac{1}{10}}=\frac{6}{10}\times \frac{10}{9}(1-\frac{1}{{10}^n})=\frac{2}{3}(1-\frac{1}{{10}^n})$$$$=\frac{2}{3}-\frac{2}{3.{10}^n}$$

Answered by $@ty@m123 last updated on 03/May/20

$$\left(1\right)1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{n-1}-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+1}$$$$=1-\frac{1}{n+1}=\frac{n}{n+1}$$$$\left(2\right)a=0.6,r=0.1$$$$\therefore S_{n-1}=\frac{a(1-r^n)}{1-r}$$$$=\frac{0.6(1-0.1^n)}{1-0.1}$$$$=\frac{0.6(1-0.1^n)}{0.9}$$$$=\frac{2(1-0.1^n)}{3}$$

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