Q-[1/3+1/50]+[1/3+2/50]+[1/3+3/50]+.......+[1/3+70/50]
what is sum of above series if[.] denote integer value

• from (1/3 +1/50) to (1/3 + 33/50) , each term is less than 1
  
  so, [1/3+1/50]+[1/3+2/50]+[1/3+3/50]+.......+[1/3+33/50] = 0
  
  from [1/3 +34/50] to [1/3 +70/50], each term is greater than 1 & have a value 1.
  from 34 to 70, no. of terms = 70-34+1 = 37
  
  ans = 1+1+...37 times = 37
• 10 years agoHelpfull: Yes(31) No(2)
• anwer is 21 as [1/3]+[1/3]...[1/3] take out common from each of the terms .71*[1/3]=71*0=0 as[0.33..]=0and next terms will continue as[1/50]to [49/50]=0 and from [50/50] to [70/50] 21 terms with value 1 so 21*1 = 21.
• 10 years agoHelpfull: Yes(2) No(5)
• Upto 33/50 the integer will be 0 and after it the integer value will be 1.. I.e.
  
  From 34 to 70
  => (70-34)*1
  => 36+1(including 34)
  
  => ans. Is 37
• 10 years agoHelpfull: Yes(2) No(1)
• this can be written as [1/3+1/3+1/3.....70terms]+[1/50+2+50....70terms]
  =[70/3]+(1/50)[1+2+3+4.....70terms]
  =[70/3]+(1/50)[2485]
  ans=73.03
• 10 years agoHelpfull: Yes(1) No(2)
• 1/3(70)+[ (1/50)+(2/50)..........(70/50)]=2191/30;
  
• 10 years agoHelpfull: Yes(1) No(1)