In a Sequence of A(n)=A(n-1)-A(n-2) where A(n) is the nTH term in the sequence n is the integer and n>=3,A(1)=1,A(2)=1 Calculate S(1000),where S(1000) is the sum of first 1000 terms?

• A(n)=A(n-1)-A(n-2)
  A(1)=1
  A(2)=1
  A(3)=A(3-1)-A(3-2)=A2-A1=1-1=0
  A(4)=A3-A2=0-1=-1
  A(5)=A4-A3=-1-0=-1
  A(6)=A5-A4=-1-(-1)=0
  A(7)=A6-A5=0-(-1)=1
  A(8)=A7-A6=1-0=1
  A(9)=A8-A7=1-1=0
  A(10)=A9-A8=0-1=-1 .... so on
  
  (1,1,0,-1,-1,0),(1,1,0,-1,...)
  we see that series is repeating after 1st 6 terms
  
  sum of 6 terms=1+1+0-1-1+0=0
  
  1000=6*166+4=996+4
  
  
  S(1000)=A(1)+A(2)+A(3)+....+A(1000)
  =sum of ist 996 terms + last 4 terms
  =(1+1+0-1-1+0)+(1+1+0-1-1+0)+...166 times +(1+1+0-1)
  = 0+0+0+...166 times + (1+1+0-1)
  =1
• 10 years agoHelpfull: Yes(26) No(0)
• answer should be 1..pateern repeats for evry six termsi.e1,1,0,-1,-1,0..sum of first 6 terms is 0..up to 996 sum becomes zero..hence frm 996 to 1000..sum equal to 1+1+0+-1=1..therefore ans is 1..thank u..
• 10 years agoHelpfull: Yes(1) No(0)
• if you solve the expression A(n)=A(n-1)-A(n-2)
  put n= 3 you get, A(3)=A(2)-A(1)=1-1=0
  thus now put n= 4... you will get series as 1,1,0,-1,-1,0,1,1,0,-1...
  here S(10)i.e. sum of 10 terms = 1
  thus sum of 1000 terms will be 1000
• 10 years agoHelpfull: Yes(0) No(2)
• f(3)=1 and f(4)=-1
  so there cancel each other
  so on........f(999)=1 and f(1000)=-1 cancel each other
  ans is 0
• 10 years agoHelpfull: Yes(0) No(0)