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Question Number 84677 by jagoll last updated on 15/Mar/20

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

Commented by Tony Lin last updated on 15/Mar/20

$$\frac{1}{1-x}=1+x+x^2+x^3+\cdot \cdot \cdot$$$$\frac{x}{1-x}=x+x^2+x^3+x^4+\cdot \cdot \cdot$$$$\frac{1}{{(1-x)}^2}=1+2x+3x^2+4x^3+\cdot \cdot \cdot$$$$\frac{x}{{(1-x)}^2}=x+2x^2+3x^3+4x^4+\cdot \cdot \cdot$$$$\frac{(1+x)}{{(1-x)}^3}=1^2+2^{2}x+3^2{x}^2+4^2{x}^3+\cdot \cdot \cdot$$$$\frac{x(1+x)}{{(1-x)}^3}=1^{2}x+2^2{x}^2+3^2{x}^3+4^2{x}^4+\cdot \cdot \cdot$$$$plug\frac{1}{2}in$$$$\Rightarrow \frac{\frac{1}{2}\times \frac{3}{2}}{\frac{1}{8}}=6$$

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