Find the sum: √(1 + 1/12 + 1/22) + √(1 + 1/22 + 1/32) + ..... + √(1 + 1/20072 + 1/20082)

a. 2008 - 1/2008

b. 2007 - 1/2007

c. 2007 - 1/2008

d. 2008 - 1/2007

• We can calculate it by the formula of arithmetic progression that is
  S=N/2[2a+(N-1)d].Therefore the correct answer is(a).
• 10 years agoHelpfull: Yes(0) No(1)
• Take the first term:
  √(1 + 1/12 + 1/22)
  = √(1 + 1 + 1/4)
  = √(9/4) = 3/2 =
  2 - 1/2
  Take the first two terms:
  √(1 + 1/12 + 1/22) + √(1 + 1/22 + 1/32)
  = 3/2 + √(1 + 1/4 + 1/9)
  = 3/2 + √(49/36)
  = 3/2 + 7/6
  = 8/3
  3 - 1/3
  Similarly:
  √(1 + 1/12 + 1/22) + √(1 + 1/22 + 1/32) + ..... + √(1 + 1/20072 + 1/20082)
  = 2008 - 1/2008
  So option (a) is correct
• 8 years agoHelpfull: Yes(0) No(1)