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Question Number 131825 by liberty last updated on 09/Feb/21

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{2\mathrm{n}-1}{2^{\mathrm{n}}}?$$

Answered by mr W last updated on 09/Feb/21

$$method1$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{x}^{2n}=\frac{x^2}{1-x^2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{x}^{2n-1}=\frac{x}{1-x^2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}(2n-1)x^{2n-2}=\frac{1}{1-x^2}+\frac{2x^2}{{(1-x^2)}^2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}(2n-1)x^{2n}=\frac{x^2}{1-x^2}+\frac{2x^4}{{(1-x^2)}^2}$$$$x^2=\frac{1}{2}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2n-1}{2^n}=\frac{\frac{1}{2}}{1-\frac{1}{2}}+\frac{2\times \frac{1}{4}}{{(1-\frac{1}{2})}^2}=1+2=3$$

Commented by liberty last updated on 09/Feb/21

$$\mathrm{cool}...$$

Commented by mr W last updated on 09/Feb/21

$$method2:$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2n-1}{2^n}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n}{2^{n-1}}-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2^n}=S_1-S_2$$$$S_2=\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^3+...=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1$$$$S_1=1+\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...$$$$\frac{1}{2}{S}_1=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...$$$$(1-\frac{1}{2})S_1=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...=1+S_2=2$$$$\Rightarrow S_1=2\times 2=4$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2n-1}{2^n}=4-1=3$$

Answered by Dwaipayan Shikari last updated on 09/Feb/21

$$Method\lfloor \pi \rfloor :)$$$$S=x+3x^2+5x^3+7x^4+..$$$$S(1-x)=x+2x^2+2x^2+...$$$$S(1-x)=\frac{2x}{(1-x)}-x\Rightarrow S=\frac{2x}{{(1-x)}^2}-\frac{x}{(1-x)}(x=\frac{1}{2})$$$$S=4-1=3$$

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