... l( a 2 ) +l( a n−3)isnot a label.(ii) By (2. 1), l( a 2 ) +l( a n 2 )=b − b − l( a n−3)= l( a n−3), then l( a 2 ) +l( a n 2 )is not a label.(iii) By (2. 1), l( a 2 ) +l( a n−1) =a +2b, then we have |l( a 4)| ... (2. 1), l( a 3) +l( a n−1) =a − b + a + b = 2a, then we have |l( a 1)| < ;l( a 3) +l( a n−1) < |l( a 3)| when b> 3a. |l( a 3)| < ;l( a 3) +l( a n−1) < |l( a 2 )| when 2a& lt;b< 3a. So l( a 3) +l( a n−1) ... our labels, l( a i) +l( a n−3) is not a label.(ii) For l( a i) +l( a n 2 ), by (2. 1) and properties (1), (2) we have |l( a i) +l( a n 2 )| = l( a i) − l( a n 2 )=b + l( a n−3) − l( a i)=b + a − l( a n−4)...