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Question-74786

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Question Number 74786 by chess1 last updated on 30/Nov/19
Commented by abdomathmax last updated on 30/Nov/19
S =Σ_(k=1) ^(2019) k((1/(2020)))^k =w((1/(2020))) with w(x)=Σ_(k=1) ^(2019) kx^k we have for x≠1 Σ_(k=0) ^(2019) x^k =((x^(2020) −1)/(x−1)) ⇒Σ_(k=1) ^(2019) kx^(k−1) =(d/dx)(((x^(2020) −1)/(x−1))) =((2020 x^(2019) (x−1)−(x^(2020) −1))/((x−1)^2 )) =((2019 x^(2020) −2020 x^(2019) +1)/((x−1)^2 )) ⇒ Σ_(k=1) ^(2019) kx^k =((2019 x^(2021) −2020 x^(2020) +x)/((x−1)^2 )) ⇒ S =((2019((1/(2020)))^(2021) −2020((1/(2020)))^(2020) +(1/(2020)))/((1−(1/(2020)))^2 ))
$${S}\:=\sum_{{k}=\mathrm{1}} ^{\mathrm{2019}} \:{k}\left(\frac{\mathrm{1}}{\mathrm{2020}}\right)^{{k}} \:={w}\left(\frac{\mathrm{1}}{\mathrm{2020}}\right)\:{with} \\ $$$${w}\left({x}\right)=\sum_{{k}=\mathrm{1}} ^{\mathrm{2019}} \:{kx}^{{k}} \:\:\:{we}\:{have}\:{for}\:{x}\neq\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{0}} ^{\mathrm{2019}} \:{x}^{{k}} \:=\frac{{x}^{\mathrm{2020}} −\mathrm{1}}{{x}−\mathrm{1}}\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{\mathrm{2019}} \:{kx}^{{k}−\mathrm{1}} \:=\frac{{d}}{{dx}}\left(\frac{{x}^{\mathrm{2020}} −\mathrm{1}}{{x}−\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{2020}\:{x}^{\mathrm{2019}} \left({x}−\mathrm{1}\right)−\left({x}^{\mathrm{2020}} −\mathrm{1}\right)}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2019}\:{x}^{\mathrm{2020}} −\mathrm{2020}\:{x}^{\mathrm{2019}} \:+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{\mathrm{2019}} \:{kx}^{{k}} \:\:=\frac{\mathrm{2019}\:{x}^{\mathrm{2021}} −\mathrm{2020}\:{x}^{\mathrm{2020}} +{x}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${S}\:=\frac{\mathrm{2019}\left(\frac{\mathrm{1}}{\mathrm{2020}}\right)^{\mathrm{2021}} −\mathrm{2020}\left(\frac{\mathrm{1}}{\mathrm{2020}}\right)^{\mathrm{2020}} +\frac{\mathrm{1}}{\mathrm{2020}}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2020}}\right)^{\mathrm{2}} } \\ $$$$ \\ $$
Answered by $@ty@m123 last updated on 01/Dec/19
Let a=(1/(2020)) S=a+2a^2 +.....+2019a^(2019) In this Arithmetico-Geometric series, First term of AP=1, common diff.=1 First term of GP=a, common rato=a ⇒S=(a/(1−a))+((a^2 (1−a^(2019) ))/((1−a)^2 ))−((2019.a^(2020) )/(1−a)) Now put a=(1/(2020)) & simplify
$${Let}\:{a}=\frac{\mathrm{1}}{\mathrm{2020}} \\ $$$${S}={a}+\mathrm{2}{a}^{\mathrm{2}} +…..+\mathrm{2019}{a}^{\mathrm{2019}} \\ $$$${In}\:{this}\:{Arithmetico}-{Geometric}\:{series}, \\ $$$${First}\:{term}\:\:{of}\:{AP}=\mathrm{1},\:{common}\:{diff}.=\mathrm{1} \\ $$$${First}\:{term}\:\:{of}\:{GP}={a},\:{common}\:{rato}={a} \\ $$$$\Rightarrow{S}=\frac{{a}}{\mathrm{1}−{a}}+\frac{{a}^{\mathrm{2}} \left(\mathrm{1}−{a}^{\mathrm{2019}} \right)}{\left(\mathrm{1}−{a}\right)^{\mathrm{2}} }−\frac{\mathrm{2019}.{a}^{\mathrm{2020}} }{\mathrm{1}−{a}} \\ $$$${Now}\:{put}\:{a}=\frac{\mathrm{1}}{\mathrm{2020}}\:\:\&\:{simplify} \\ $$
Answered by mr W last updated on 01/Dec/19
1+x+x^2 +...+x^n =((1−x^(n+1) )/(1−x)) 1+2x+3x^2 ...+nx^(n−1) =((1−x^(n+1) )/((1−x)^2 ))−(((n+1)x^n )/(1−x)) x+2x^2 +3x^3 ...+nx^n =((x(1−x^(n+1) ))/((1−x)^2 ))−(((n+1)x^(n+1) )/(1−x)) put x=(1/(2020)), n=2019 we get (1/(2020))+(2/(2020^2 ))+(3/(2020^3 ))+...+((2019)/(2020^(2019) ))=((1−(1/(2020^(2020) )))/(2020(1−(1/(2020)))^2 ))−((2020)/((1−(1/(2020)))2020^(2020) )) (1/(2020))+(2/(2020^2 ))+(3/(2020^3 ))+...+((2019)/(2020^(2019) ))=((2020^(2020) −1)/(2019^2 ×2020^(2019) ))−(1/(2019×2020^(2018) )) ⇒(1/(2020))+(2/(2020^2 ))+(3/(2020^3 ))+...+((2019)/(2020^(2019) ))=((2020(2020^(2019) −2019)−1)/(2019^2 ×2020^(2019) ))
$$\mathrm{1}+{x}+{x}^{\mathrm{2}} +…+{x}^{{n}} =\frac{\mathrm{1}−{x}^{{n}+\mathrm{1}} }{\mathrm{1}−{x}} \\ $$$$\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} …+{nx}^{{n}−\mathrm{1}} =\frac{\mathrm{1}−{x}^{{n}+\mathrm{1}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }−\frac{\left({n}+\mathrm{1}\right){x}^{{n}} }{\mathrm{1}−{x}} \\ $$$${x}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} …+{nx}^{{n}} =\frac{{x}\left(\mathrm{1}−{x}^{{n}+\mathrm{1}} \right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }−\frac{\left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{1}} }{\mathrm{1}−{x}} \\ $$$${put}\:{x}=\frac{\mathrm{1}}{\mathrm{2020}},\:{n}=\mathrm{2019}\:{we}\:{get} \\ $$$$\frac{\mathrm{1}}{\mathrm{2020}}+\frac{\mathrm{2}}{\mathrm{2020}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2020}^{\mathrm{3}} }+…+\frac{\mathrm{2019}}{\mathrm{2020}^{\mathrm{2019}} }=\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2020}^{\mathrm{2020}} }}{\mathrm{2020}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2020}}\right)^{\mathrm{2}} }−\frac{\mathrm{2020}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2020}}\right)\mathrm{2020}^{\mathrm{2020}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2020}}+\frac{\mathrm{2}}{\mathrm{2020}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2020}^{\mathrm{3}} }+…+\frac{\mathrm{2019}}{\mathrm{2020}^{\mathrm{2019}} }=\frac{\mathrm{2020}^{\mathrm{2020}} −\mathrm{1}}{\mathrm{2019}^{\mathrm{2}} ×\mathrm{2020}^{\mathrm{2019}} }−\frac{\mathrm{1}}{\mathrm{2019}×\mathrm{2020}^{\mathrm{2018}} } \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2020}}+\frac{\mathrm{2}}{\mathrm{2020}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2020}^{\mathrm{3}} }+…+\frac{\mathrm{2019}}{\mathrm{2020}^{\mathrm{2019}} }=\frac{\mathrm{2020}\left(\mathrm{2020}^{\mathrm{2019}} −\mathrm{2019}\right)−\mathrm{1}}{\mathrm{2019}^{\mathrm{2}} ×\mathrm{2020}^{\mathrm{2019}} } \\ $$
Commented by chess1 last updated on 01/Dec/19
thanks
$$\mathrm{thanks} \\ $$

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