find sum of 3+5/(1+2^2)+7/(1+2^2+3^2)+.......

• this is in da form of summation of terms each in form of [(2n+1)/(1^1+2^2+...n^n)]
  where n is from 1 to infinity
  1^1+2^2+...n^n=[n*(n+1)*(2n+1)]/6
  so,(2n+1)/(1^1+2^2+...n^n)=sigma of [6/(n*(n+1))]
  =6[(1/(2*3))+(1/(2*3))....]
  =6[(1/1-1/2)+(1/2-1/3)+....]
  =6
  therefore ans is 6
  
• 10 years agoHelpfull: Yes(7) No(2)
• 6
  do the partial summation
  
• 10 years agoHelpfull: Yes(2) No(6)
• 3+1+0.5+0.25+0.125+0.0750+0.03750+..........
  =5 approximate
• 10 years agoHelpfull: Yes(1) No(2)
• continuation of SOUJANYA KOLUKULURI
  => 6[1-1/n]
  
  without knowing n value how it is get cancelled please explain
• 7 years agoHelpfull: Yes(1) No(0)