Question Number 136161 by nimnim last updated on 19/Mar/21
$${How}\:{do}\:{I}\:{find}\:{the}\:{sum}\:{of} \\ $$$$\:\:\mathrm{1}+\mathrm{3}{x}+\mathrm{6}{x}^{\mathrm{2}} +\mathrm{10}{x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{4}} +…….\infty,\:\:{where}\:−\mathrm{1}<{x}<\mathrm{1}\:? \\ $$$$\:\:{Please}\:{Help}.. \\ $$
Answered by bramlexs22 last updated on 19/Mar/21
$${S}\:=\:\mathrm{1}+\mathrm{3}{x}+\mathrm{6}{x}^{\mathrm{2}} +\mathrm{10}{x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{4}} +… \\ $$$${xS}={x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}{x}^{\mathrm{3}} +\mathrm{10}{x}^{\mathrm{4}} +… \\ $$$$\left(\mathrm{1}−{x}\right){S}=\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{4}} +… \\ $$$${let}\:{T}\:=\:\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{4}} +… \\ $$$$\int\:{T}\:{dx}\:=\:{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +{x}^{\mathrm{5}} +… \\ $$$$\int{T}\:{dx}\:=\:\frac{{x}}{\mathrm{1}−{x}}\:{then}\:{T}=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$${so}\:{S}\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} } \\ $$
Commented by nimnim last updated on 19/Mar/21
$${Thank}\:{you}\:{Sir}. \\ $$