Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 83638 by bshahid010@gmail.com last updated on 04/Mar/20

Answered by TANMAY PANACEA last updated on 05/Mar/20

S=(1/(x+1))+(2/(x^2 +1))+(4/(x^4 +1))+...+(2^n /(x^2^n +1)) (i think) S=Σ_(n=0) ^n (2^n /(x^2^n +1)) total (n+1)terms S−(1/(x−1))=((1/(x+1))−(1/(x−1)))+(2/(x^2 +1))+(4/(x^4 +1))+..+(2^n /(x^2^n +1)) S−(1/(x−1))=(((−2)/(x^2 −1))+(2/(x^2 +1)))+(4/(x^4 +1))+..+(2^n /(x^2^n +1)) s−(1/(x−1))=(((−4)/(x^4 −1))+(4/(x^4 +1)))+..+((2^n /(x^2^n +1))) S−(1/(x−1))=(((−8)/(x^8 −1)))+...+((2^n /(x^2^n +1))) ..... ...... S−(1/(x−1))=(((−2^(n+1) )/(x^2^(n+1) −1))) S=(1/(x−1))−(2^(n+1) /(x^2^(n+1) −1))

S=1x+1+2x2+1+4x4+1+...+2nx2n+1(ithink)S=∑nn=02nx2n+1total(n+1)termsS−1x−1=(1x+1−1x−1)+2x2+1+4x4+1+..+2nx2n+1S−1x−1=(−2x2−1+2x2+1)+4x4+1+..+2nx2n+1s−1x−1=(−4x4−1+4x4+1)+..+(2nx2n+1)S−1x−1=(−8x8−1)+...+(2nx2n+1)...........S−1x−1=(−2n+1x2n+1−1)S=1x−1−2n+1x2n+1−1

Terms of Service

Privacy Policy

Contact: info@tinkutara.com