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

Σ_(k = 1) ^∞ (k^2 /2^k ) ?

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

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

(1/(1−x))=1+x+x^2 +x^3 +∙∙∙ (x/(1−x))=x+x^2 +x^3 +x^4 +∙∙∙ (1/((1−x)^2 ))=1+2x+3x^2 +4x^3 +∙∙∙ (x/((1−x)^2 ))=x+2x^2 +3x^3 +4x^4 +∙∙∙ (((1+x))/((1−x)^3 )) =1^2 +2^2 x+3^2 x^2 +4^2 x^3 +∙∙∙ ((x(1+x))/((1−x)^(3 ) ))=1^2 x+2^2 x^2 +3^2 x^3 +4^2 x^4 +∙∙∙ plug (1/2) in ⇒(((1/2)×(3/2))/(1/8))=6

$$\frac{\mathrm{1}}{\mathrm{1}−{x}}=\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +\centerdot\centerdot\centerdot \\ $$$$\frac{{x}}{\mathrm{1}−{x}}={x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +\centerdot\centerdot\centerdot \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }=\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} +\centerdot\centerdot\centerdot \\ $$$$\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }={x}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{4}} +\centerdot\centerdot\centerdot \\ $$$$\frac{\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} \:\:}\:=\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} {x}+\mathrm{3}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} {x}^{\mathrm{3}} +\centerdot\centerdot\centerdot \\ $$$$\frac{{x}\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}\:\:} }=\mathrm{1}^{\mathrm{2}} {x}+\mathrm{2}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} {x}^{\mathrm{3}} +\mathrm{4}^{\mathrm{2}} {x}^{\mathrm{4}} +\centerdot\centerdot\centerdot \\ $$$${plug}\:\frac{\mathrm{1}}{\mathrm{2}}\:{in} \\ $$$$\Rightarrow\frac{\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{8}}}=\mathrm{6} \\ $$

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