In a sequence of integers, A(n)=A(n-1)-A(n-2), where A(n) is the nth term in the sequence. n is an integer and n>=3, A1=-1,A2=1. Calculate S(1000) where S(1000) is the sum of the first 1000 terms

• Given A(1)=1;A(2)=1
  So, A(3)=0
  A(4)=A(3)-A(2)= -1
  A(5)=A(4)-A(3)= -1
  A(6)=A(5)-A(4)= 0
  A(7)=A(6)-A(5)= 1 = A(1)
  A(8)=A(7)-A(6)= 1 = A(2)
  So,next terms can easily be predicted from the previous steps.
  A(9)=A(3)=0
  A(10)=A(4)= -1
  A(11)=A(5)= -1
  A(12)=A(6)= 0
  A(13)=A(7)=A(1)=1
  A(14)=A(8)=A(2)=1
  So, Observing the previous steps little bit carefully , we infer that the 1st 6 terms are repeated continuously.
  Now, the sum of 1st 6 terms = A(1)+A(2)+A(3)+A(4)+A(5)+A(6)=1+1+0-1-1+0=0
  Now, 1000 = 6*166 + 4.
  so, in the 1st 1000 terms of the given sequence ,the 1st 6 terms will be repeated 166 times giving a total sum =0, and then the result will be the sum of remaining 4 terms which are effectively the 1st 4 terms of the sequence, A(1), A(2),A(3) & A(4).
  So, S(1000) = 166*0 +A(1)+A(2)+A(3)+A(4) = 0 + 1 + 1 + 0 - 1 = 1.
  
  Hence S(1000) = 1.
• 11 years agoHelpfull: Yes(11) No(4)
• s(1000)=a1+a2+a3+,,,,,+a999+a1000
  =a1+a2+a2-a1+a3-a2+a4-a3,,,,,,,,a998-a997+a999-a998
  =a2+a999
  a1=-1,a2=1,a3=2,a4=1,a5=-1,a6=-2,a7=-1,a8=1,a9=2,a10=1,a11=-1,a12=-2,a13=-1,,,,,
  the above has period of 6 (-1,1,2,1,-1,-2) repeating so
  remainde of a999=999/6= 3 so it have same value of a3=2
  s(1000)=a2+a999=1+2=3
• 11 years agoHelpfull: Yes(8) No(10)
• question has a1 value -1
• 11 years agoHelpfull: Yes(0) No(6)