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Question Number 17102 by tawa tawa last updated on 30/Jun/17

$$\mathrm{compute}:\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{2\mathrm{k}+1}{2^{2(\mathrm{k}+1)}}$$

Commented by prakash jain last updated on 01/Jul/17

$$\mathrm{S}=\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{2\mathrm{k}+1}{2^{2(\mathrm{k}+1)}}$$$${\mathrm{S}}_1=\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{2\mathrm{k}}{2^{2(\mathrm{k}+1)}}$$$${\mathrm{S}}_2=\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{2^{2(k+1)}}=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...$$$$S=S_1+S_2$$$$S_2\mathrm{is}\mathrm{infinite}\mathrm{GP}{S}_2=\frac{\frac{1}{2^2}}{1-\frac{1}{2^2}}=\frac{1}{3}$$$${\mathrm{S}}_1=\frac{2}{2^2}+\frac{3}{2^4}+\frac{4}{2^6}+...$$$$\frac{{\mathrm{S}}_1}{2^2}=\frac{2}{2^4}+\frac{3}{2^6}+...$$$${\mathrm{S}}_1-\frac{{\mathrm{S}}_1}{4}=\frac{2}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...$$$${\mathrm{S}}_1-\frac{{\mathrm{S}}_1}{4}=\frac{1}{2^2}+(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...)$$$$\frac{3}{4}{\mathrm{S}}_1=\frac{1}{2^2}+\frac{\frac{1}{2^2}}{1-\frac{1}{2^2}}=\frac{1}{4}+\frac{1}{3}=\frac{7}{12}$$$${\mathrm{S}}_1=\frac{7}{9}$$$$\mathrm{S}={\mathrm{S}}_1+{\mathrm{S}}_2=\frac{7}{9}+\frac{1}{3}=\frac{10}{9}$$

Commented by ajfour last updated on 01/Jul/17

$$\mathrm{utter}\mathrm{simple}\mathrm{for}\mathrm{you}\mathrm{Sir}.$$

Commented by tawa tawa last updated on 01/Jul/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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