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Question-120160

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Question Number 120160 by 77731 last updated on 29/Oct/20
Commented by JDamian last updated on 29/Oct/20
https://m.youtube.com/watch?v=9kfmk7jEuvQ
Answered by mindispower last updated on 29/Oct/20
(1/(2^m (2^n +2^m )))=−(1/(2^n (2^n +2^m )))+(1/2^(n+m) ) Σ_(m≥0) Σ_(n≥0) ((m+n+nm)/(2^m (2^n +2^m )))=−Σ_(m≥0) Σ_(n≥0) ((m+n+nm)/(2^n (2^n +2^m )))+ΣΣ((n+m+nm)/2^(n+m) ) S=Σ_(m≥0) Σ_(n≥0) ((m+n+nm)/(2^m (2^n +2^m )))=Σ_m Σ_n ((n+m+nm)/(2^n (2^m +2^n )))=S (n,m)→(m,n) is bijection ⇔ S=−S+2Σ_m Σ_n (m/2^(n+m) )+ΣΣ((mn)/(2^n .2^m )) Σ_(m≥0) (m/2^m )=Σ_(m≥1) (m/2^m ),(1/(1−x))=Σx^n ⇒(x/((1−x)^2 ))=Σ_(n≥1) nx^n 2S=2.(1/(1−(1/2))).((1/2)/((1−(1/2))^2 ))+(((1/2)/((1−(1/2))^2 )))^2 2S=4.2+4⇒S=6 Σ_(n≥0) Σ_(m≥0) ((n+m+nm)/(2^m (2^m +2^n )))=6
$$\frac{\mathrm{1}}{\mathrm{2}^{{m}} \left(\mathrm{2}^{{n}} +\mathrm{2}^{{m}} \right)}=−\frac{\mathrm{1}}{\mathrm{2}^{{n}} \left(\mathrm{2}^{{n}} +\mathrm{2}^{{m}} \right)}+\frac{\mathrm{1}}{\mathrm{2}^{{n}+{m}} } \\ $$$$\underset{{m}\geqslant\mathrm{0}} {\sum}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{{m}+{n}+{nm}}{\mathrm{2}^{{m}} \left(\mathrm{2}^{{n}} +\mathrm{2}^{{m}} \right)}=−\underset{{m}\geqslant\mathrm{0}} {\sum}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{{m}+{n}+{nm}}{\mathrm{2}^{{n}} \left(\mathrm{2}^{{n}} +\mathrm{2}^{{m}} \right)}+\Sigma\Sigma\frac{{n}+{m}+{nm}}{\mathrm{2}^{{n}+{m}} } \\ $$$${S}=\underset{{m}\geqslant\mathrm{0}} {\sum}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{{m}+{n}+{nm}}{\mathrm{2}^{{m}} \left(\mathrm{2}^{{n}} +\mathrm{2}^{{m}} \right)}=\underset{{m}} {\sum}\underset{{n}} {\sum}\frac{{n}+{m}+{nm}}{\mathrm{2}^{{n}} \left(\mathrm{2}^{{m}} +\mathrm{2}^{{n}} \right)}={S} \\ $$$$\left({n},{m}\right)\rightarrow\left({m},{n}\right)\:{is}\:{bijection} \\ $$$$\Leftrightarrow \\ $$$${S}=−{S}+\mathrm{2}\underset{{m}} {\sum}\underset{{n}} {\sum}\frac{{m}}{\mathrm{2}^{{n}+{m}} }+\Sigma\Sigma\frac{{mn}}{\mathrm{2}^{{n}} .\mathrm{2}^{{m}} } \\ $$$$\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{{m}}{\mathrm{2}^{{m}} }=\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{{m}}{\mathrm{2}^{{m}} },\frac{\mathrm{1}}{\mathrm{1}−{x}}=\Sigma{x}^{{n}} \Rightarrow\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }=\underset{{n}\geqslant\mathrm{1}} {\sum}{nx}^{{n}} \\ $$$$\mathrm{2}{S}=\mathrm{2}.\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}.\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }+\left(\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{S}=\mathrm{4}.\mathrm{2}+\mathrm{4}\Rightarrow{S}=\mathrm{6} \\ $$$$\underset{{n}\geqslant\mathrm{0}} {\sum}\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{{n}+{m}+{nm}}{\mathrm{2}^{{m}} \left(\mathrm{2}^{{m}} +\mathrm{2}^{{n}} \right)}=\mathrm{6} \\ $$
Commented by 77731 last updated on 01/Nov/20
very nice work ...!!

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