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How-do-I-find-the-sum-of-1-3x-6x-2-10x-3-15x-4-where-1-lt-x-lt-1-Please-Help-




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

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