What is the value of 1/1 + 1/3 + 1/6 + 1/10 + ...

a. 2

b. 3/2

c. 5/3

d. 3

• a. 2
  
  Tn = 1/[n(n+1)/2] = 2/n(n+1) = 2 *[1/n -1/(n+1)]
  
  Sum = 1/1 + 1/3 + 1/6 + 1/10 + ...
  = 2[(1 -1/2)+ (1/2 -1/3) + (1/3 -1/4) +...+ (1/n -1/(n+1) ]
  = 2[1 - 1/(n+1)]
  
  as n tends infinity 1/(n+1) goes to 0.
  
  S = 2*(1-0) = 2
• 10 years agoHelpfull: Yes(31) No(0)
• 1/1 + 1/3 + 1/6 + 1/10
  
  = 2/1*2 + 2/2*3 + 2/3*4 + 2/4*5 + ....
  
  = 2(1-1/2) + 2(1/2 - 1/3) + 2(1/3 - 1/4) + 2(1/4 - 1/5) + ....
  
  = 2[1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ....]
  
  = 2(1)
  
  = 2
• 10 years agoHelpfull: Yes(21) No(0)
• ans is 3....as upto 7th term sum is equal to 2.1...so for greater terms it will b more than 2.1....which could be only 3 as per the options are considered
  
• 10 years agoHelpfull: Yes(1) No(13)
• bl fifa khelbi uthe porbe
• 10 years agoHelpfull: Yes(0) No(0)
• 1+o.33+0.166+0.1+0.06+0.04+0.04+..now smaller values..=>>>1.72
  so,near 2
• 10 years agoHelpfull: Yes(0) No(0)