In a sequence of integer A(n)=A(n-1)-A(n-2)where A(n) is nth term, n is integer and n>=3,A(1)=1,A(2)=1,calculate S(1000)

• 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
  
  
  we see that series is repeating after 1st 6 terms
  
  sum of 1st 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
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