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1-6-1-2-6-2-3-6-3-n-6-n-

Question Number 125663 by Mammadli last updated on 12/Dec/20 $$\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{1}} }+\frac{\mathrm{2}}{\mathrm{6}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{6}^{\mathrm{3}} }+…+\frac{{n}}{\mathrm{6}^{{n}} }=? \\ $$ Answered by Dwaipayan Shikari last updated on 12/Dec/20 $${S}=\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{2}}{\mathrm{6}^{\mathrm{2}}…

a-x-b-y-c-z-x-a-y-b-z-c-

Question Number 125661 by Mammadli last updated on 12/Dec/20 $$\boldsymbol{{a}}^{\boldsymbol{{x}}} \:=\:\boldsymbol{{b}}^{\boldsymbol{{y}}} \:=\:\boldsymbol{{c}}^{\boldsymbol{{z}}} \\ $$$$\boldsymbol{{x}}^{\boldsymbol{{a}}} \:+\:\boldsymbol{{y}}^{\boldsymbol{{b}}} \:+\:\boldsymbol{{z}}^{\boldsymbol{{c}}} \:=\:? \\ $$ Terms of Service Privacy Policy Contact:…

1-2020-1-2-2020-2-3-2020-3-n-2020-n-

Question Number 125659 by Mammadli last updated on 12/Dec/20 $$\frac{\mathrm{1}}{\mathrm{2020}^{\mathrm{1}} }+\frac{\mathrm{2}}{\mathrm{2020}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2020}^{\mathrm{3}} }+…+\frac{\boldsymbol{{n}}}{\mathrm{2020}^{\boldsymbol{{n}}} }=? \\ $$ Answered by Dwaipayan Shikari last updated on 12/Dec/20 $${S}=\frac{\mathrm{1}}{\mathrm{2020}}+\frac{\mathrm{2}}{\mathrm{2020}^{\mathrm{2}}…

Question-125651

Question Number 125651 by Algoritm last updated on 12/Dec/20 Answered by Olaf last updated on 12/Dec/20 $$\begin{cases}{{y}_{\mathrm{1}} '\:=\:{y}_{\mathrm{1}} +\mathrm{4}{y}_{\mathrm{2}} \:\left(\mathrm{1}\right)}\\{{y}_{\mathrm{2}} '\:=\:−{y}_{\mathrm{1}} +{y}_{\mathrm{2}} +{e}^{\mathrm{3}{x}} \:\left(\mathrm{2}\right)}\end{cases} \\…

In-a-certain-set-of-200-students-subset-A-are-blondes-and-the-rest-are-brunettes-B-are-left-handed-and-C-are-left-handed-blondes-1-Express-in-terms-of-A-B-C-the-subset-consisting-of-right-han

Question Number 125645 by aurpeyz last updated on 12/Dec/20 $${In}\:{a}\:{certain}\:{set}\:{of}\:\mathrm{200}\:{students}.\: \\ $$$${subset}\:{A}\:{are}\:{blondes}\:{and}\:{the}\:{rest}\:{are} \\ $$$${brunettes}.\:{B}\:{are}\:{left}\:{handed}.\:{and}\: \\ $$$${C}\:{are}\:{left}−{handed}\:{blondes}. \\ $$$$ \\ $$$$\left(\mathrm{1}\right){Express}\:{in}\:{terms}\:{of}\:{A}\:{B}\:{C}\:{the}\: \\ $$$${subset}\:{consisting}\:{of}\:{right}−{handed} \\ $$$${brunettes}. \\…

Question-125638

Question Number 125638 by aurpeyz last updated on 12/Dec/20 Answered by mathmax by abdo last updated on 12/Dec/20 $$\mathrm{S}\:=\frac{\mathrm{3}}{\mathrm{3}}−\frac{\mathrm{5}}{\mathrm{5}^{\mathrm{2}} }+\frac{\mathrm{7}}{\mathrm{5}^{\mathrm{3}} }−\frac{\mathrm{9}}{\mathrm{5}^{\mathrm{4}} }+…\Rightarrow\mathrm{S}\:=\mathrm{1}+\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{a}_{\mathrm{n}}…

Question-125641

Question Number 125641 by aurpeyz last updated on 12/Dec/20 Answered by Dwaipayan Shikari last updated on 12/Dec/20 $${S}=\mathrm{2}+\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{8}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{11}}{\mathrm{2}^{\mathrm{3}} }+… \\ $$$$\frac{{S}}{\mathrm{2}}=\:\:\:\:\:\:\frac{\mathrm{2}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{8}}{\mathrm{2}^{\mathrm{3}} }+.. \\…