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Question Number 184927 by mathlove last updated on 14/Jan/23

$$s=\frac{2}{{11}^0}+\frac{6}{11}+\frac{10}{{11}^2}+\frac{14}{{11}^3}+\frac{18}{{11}^4}+\cdot \cdot \cdot \cdot$$$$s=?$$

Answered by Rasheed.Sindhi last updated on 14/Jan/23

$$s=\frac{2}{{11}^0}+\frac{6}{11}+\frac{10}{{11}^2}+\frac{14}{{11}^3}+\frac{18}{{11}^4}+\cdot \cdot \cdot \cdot$$$$\frac{s}{11}=\frac{2}{{11}^1}+\frac{6}{{11}^2}+\frac{10}{{11}^3}+\frac{14}{{11}^4}+\frac{18}{{11}^5}+\cdot \cdot \cdot \cdot$$$$s-\frac{s}{11}=\frac{2}{{11}^0}+\frac{4}{{11}^{}}+\frac{4}{{11}^2}+\frac{4}{{11}^3}+...$$$$\frac{10s}{11}=2+4(\frac{1}{{11}^{}}+\frac{1}{{11}^2}+\frac{1}{{11}^3}+...)$$$$10s=22+44\left(\frac{1/11}{1-1/11}\right)$$$$10s=22+44(\frac{1}{11}\cdot \frac{11}{10})=22+\frac{22}{5}=\frac{132}{5}$$$$s=\frac{132}{10\times 5}=\frac{66}{25}$$

Commented by mathlove last updated on 14/Jan/23

$$thanksdear$$

Answered by Rasheed.Sindhi last updated on 14/Jan/23

$$\mathbb{A}\mathbf{n}\mathbb{O}\mathbf{ther}\mathbb{W}\mathbf{ay}...$$$$\left(i\right)\bullet s=\frac{2}{{11}^0}+\frac{6}{11}+\frac{10}{{11}^2}+\frac{14}{{11}^3}+\frac{18}{{11}^4}+\cdot \cdot \cdot \cdot$$$$s-2=\frac{1}{11}(\frac{6}{{11}^0}+\frac{10}{{11}^1}+\frac{14}{{11}^2}+\frac{18}{{11}^3}+\cdot \cdot \cdot \cdot$$$$\left(ii\right)\bullet 11(s-2)=\frac{6}{{11}^0}+\frac{10}{{11}^1}+\frac{14}{{11}^2}+\frac{18}{{11}^3}+\cdot \cdot \cdot \cdot$$$$\left(ii\right)-\left(i\right):$$$$11(s-2)-s=\frac{4}{{11}^0}+\frac{4}{{11}^1}+\frac{4}{{11}^2}+\frac{4}{{11}^3}+...$$$$\frac{10s-22}{4}=1+\frac{1}{11}+\frac{1}{{11}^2}+\frac{1}{{11}^3}+...$$$$=\frac{1}{1-\frac{1}{11}}=\frac{11}{10}$$$$10s-22=\frac{44}{10}$$$$10s=\frac{22}{5}+22=\frac{132}{5}$$$$s=\frac{132}{50}=\frac{66}{25}$$

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