A sequence of integers `a_1+a_2++a_n` satisfies `a_(n+2)=a_(n+1)-a_nforngeq1` . Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is -99. Find the sum of the first 2002 terms.

Explanation

For every integer,`nge1`. We have
`a_(n+2)=a_(n+1)-a_(n)`
`thereforea_(n+3)=a_(n+2)-a_(n+1)`
`=a_(n+1)-a_(n)-a_(n+1)`
`=-a_(n)`
`a_(n+4)=a_(n+3)-a_(n+2)`
`a_(n+5)=a_(n+4)-a_(n+3)`
`therefore a_(n+4)+a_(n+5)=a_(n+4)-a_(n+2)`
`therefore a_(n+5)=-a_(n+2)`
`thereforea_(n)+a_(n+1)+a_(n+2)+a_(n+3)+a_(n+4)+a_(n+5)`
`=a_(n)+a_(n+1)+a_(n+2)-a_(n)+a_(n+3)-a_(n+2)-a_(n+2)`
`=a_(n+1)+a_(n+3)-a_(n+2)=0`
Thus, the sum of every six consecutive terms is 0.
Let `S_(n)` denotes the sum of first n terms
`S_(999)=S_(6xx166+3)=S_(3)" " (because ` every'6' consecutive terms has sum zero.)
`S_(1003)=S_(6xx167+1)=S_(1)`
`S_(2002)=S_(6xx333+4)=S_(4)`
`S_(2002)=S_(4)=a_(1)+a_(2)+a_(3)+a_(4)`
`=S_(3)-a_(1) (because_(4)=-a_(1) " since " a_(n+3)=-a_(n))`
=1003-(-999)
=2002