.If A(n)=A(n-1)-A(n-2), n>=3,A(1)=1, A(2)=1. Then S(1000)=?

• given that a(1)=1,a(2)=1.a(3)=0,a(4)=-1,a(5)=-1,a(6)=0,a(7)=1,a(8)=1,a(9)=0.
  so we can see that in the first three digit it is 1,1,0.then next three will be -1,-1,0.then again 1,1,0 and it is repeated in this fashion,.So upto a(999) it is repeated as it is with taking three in a pair..we can see that upto 999th term the sum is 0 bcoz 2-2+2-2+..... as the sum of first three pair is 2 and next three pair is -2.So upto 999th term it is continued and the 1000th term we get as 1..So S(1000)=1
• 11 years agoHelpfull: Yes(31) No(2)
• a(1)=1,a(2)=1,a(3)=0,a(4)=-1,a(5)=-1,a(6)=0,a(7)=1,a(8)=1,a(9)=0...for every 3rd term we are getting value as 0..and if it continues...a(999) will be 0 and a(1000) will be -1...so s(1000)=-1 and it is not 1
• 11 years agoHelpfull: Yes(9) No(3)
• s(1000)=1;
• 11 years agoHelpfull: Yes(2) No(3)
• on cal. till a(12) we get series as a(1)=1, a(2)=1,a(3)=0,a(4)=-1, a(5)=-1,a(6)=0,a(7)=1, a(8)=1,a(9)=0,a(10)=-1,a(11)=-1,a(12)=0.on summing first 6 terms we get 0 till 996th term .since 1000 is not the perfectly divisible by 6 we get remainder 4 on dividing 1000%6 which means we have to sum first 4 members of series i.e. 1+1+0-1="1"
• 11 years agoHelpfull: Yes(1) No(0)
• s(1000)=1;
  
• 11 years agoHelpfull: Yes(0) No(1)
• 1
• 11 years agoHelpfull: Yes(0) No(1)
• defntly answr is not 1 ,, its -1 ans
  seee a(1)=1,a(2)=1.a(3)=0,a(4)=-1,a(5)=-1,a(6)=0,a(7)=1,a(8)=1,a(9)=0
  here for 6 intgrs 1 cycle(1.e 1 1 0 -1 -1 0) compltes
  so 1000/6=166 (166*6=996 =>> remaing 4), remaing dose 4 shuolud be done in a seques of 1 1 0 -1
  so anser is -1 for s(1000)..
• 10 years agoHelpfull: Yes(0) No(0)
• from given A(1)=1
  A(2)=1
  A(3)=A(2)-A(1)
  A(4)=A(3)-A(2)
  A(5)=A(4)-A(3)
  .
  .
  .
  A(1000)=A(999)-A(998)
  sooo finally S(1000)=1
• 10 years agoHelpfull: Yes(0) No(0)