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Question Number 125659 by Mammadli last updated on 12/Dec/20

(1/(2020^1 ))+(2/(2020^2 ))+(3/(2020^3 ))+...+(n/(2020^n ))=?

120201+220202+320203+...+n2020n=?

Answered by Dwaipayan Shikari last updated on 12/Dec/20

S=(1/(2020))+(2/(2020^2 ))+...+(n/(2020^n )) (S/(2020))= (1/(2020^2 ))+(2/(2020^3 ))+...+((n−1)/(2020^n ))+(n/(2020^(n+1) )) S−(S/(2020))=(1/(2020))+(1/(2020^2 ))+(1/(2020^3 ))+..(1/(2020^n ))−(n/(2020^(n+1) )) ((2019S)/(2020))=((1−(2020)^(−n) )/(2019))−(n/(2020^(n+1) )) S=((2020−(2020)^(1−n) )/(2019^2 ))−(n/(2019.2020^n ))

S=12020+220202+...+n2020nS2020=120202+220203+...+n−12020n+n2020n+1S−S2020=12020+120202+120203+..12020n−n2020n+12019S2020=1−(2020)−n2019−n2020n+1S=2020−(2020)1−n20192−n2019.2020n

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