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 153847 by mathdanisur last updated on 11/Sep/21

S = x + 2x^2 + ... + nx^n

S=x+2x2+...+nxn

Answered by liberty last updated on 11/Sep/21

S=x+2x^2 +3x^3 +4x^4 +…+nx^n xS= x^2 +2x^3 +3x^4 +…+(n−1)x^n +nx^(n+1) (1−x)S=x+x^2 +x^3 +x^4 +…+x^n −nx^(n+1) (1−x)S=((x(1−x^n ))/(1−x))−nx^(n+1) S=((x(1−x^n ))/((1−x)^2 ))−((nx^(n+1) )/((1−x)))

S=x+2x2+3x3+4x4+…+nxnxS=x2+2x3+3x4+…+(n−1)xn+nxn+1(1−x)S=x+x2+x3+x4+…+xn−nxn+1(1−x)S=x(1−xn)1−x−nxn+1S=x(1−xn)(1−x)2−nxn+1(1−x)

Commented by mathdanisur last updated on 11/Sep/21

thanks ser nice

thankssernice

Commented by peter frank last updated on 11/Sep/21

thanks

thanks

Answered by mr W last updated on 11/Sep/21

Method 1: x+x^2 +...+x^n =((x^n −1)/(x−1)) 1+2x+...+nx^(n−1) =((nx^(n−1) )/(x−1))−((x^n −1)/((x−1)^2 )) x+2x^2 +...+nx^n =((nx^n )/(x−1))−(((x^n −1)x)/((x−1)^2 )) Method 2: S=x+2x^2 +3x^3 +...+nx^n xS= x^2 +2x^3 +...+(n−1)x^n +nx^(n+1) (1−x)S=x+x^2 +x^3 +...x^n −nx^(n+1) (1−x)S=(((x^n −1)x)/(x−1))−nx^(n+1) ⇒S=((nx^(n+1) )/(x−1))−(((x^n −1)x)/((x−1)^2 ))

Method1:x+x2+...+xn=xn−1x−11+2x+...+nxn−1=nxn−1x−1−xn−1(x−1)2x+2x2+...+nxn=nxnx−1−(xn−1)x(x−1)2Method2:S=x+2x2+3x3+...+nxnxS=x2+2x3+...+(n−1)xn+nxn+1(1−x)S=x+x2+x3+...xn−nxn+1(1−x)S=(xn−1)xx−1−nxn+1⇒S=nxn+1x−1−(xn−1)x(x−1)2

Commented by mathdanisur last updated on 11/Sep/21

Very nice thanks ser

Verynicethanksser

Terms of Service

Privacy Policy

Contact: info@tinkutara.com