find the sum of the series 1/40 + 1/88 + 1/154..........+ 1/59290

1) 16/245
2) 3/49
3) 48/245
4) 48/240

• 16/245
  
  1/5*8 + 1/8*11 + 1/11*14
  
  = sum (1/n(n+3)) = Sum [1/3*(1/n - 1/n+3)]
  
  = 1/3(1/5 -1/8+ 1/8 - 1/11+ 1/11 + .... - 1/245)= 1/3*( 1/5-1/245)
  
  = 1/3 * 48/245 = 16/245
• 11 years agoHelpfull: Yes(9) No(0)