Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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=∑k=12019k(12020)k=w(12020)withw(x)=∑k=12019kxkwehaveforx≠1∑k=02019xk=x2020−1x−1⇒∑k=12019kxk−1=ddx(x2020−1x−1)=2020x2019(x−1)−(x2020−1)(x−1)2=2019x2020−2020x2019+1(x−1)2⇒∑k=12019kxk=2019x2021−2020x2020+x(x−1)2⇒S=2019(12020)2021−2020(12020)2020+12020(1−12020)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

Leta=12020S=a+2a2+.....+2019a2019InthisArithmetico−Geometricseries,FirsttermofAP=1,commondiff.=1FirsttermofGP=a,commonrato=a⇒S=a1−a+a2(1−a2019)(1−a)2−2019.a20201−aNowputa=12020&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) ))

1+x+x2+...+xn=1−xn+11−x1+2x+3x2...+nxn−1=1−xn+1(1−x)2−(n+1)xn1−xx+2x2+3x3...+nxn=x(1−xn+1)(1−x)2−(n+1)xn+11−xputx=12020,n=2019weget12020+220202+320203+...+201920202019=1−1202020202020(1−12020)2−2020(1−12020)2020202012020+220202+320203+...+201920202019=20202020−120192×20202019−12019×20202018⇒12020+220202+320203+...+201920202019=2020(20202019−2019)−120192×20202019

Commented by chess1 last updated on 01/Dec/19

thanks

thanks

Terms of Service

Privacy Policy

Contact: info@tinkutara.com