记 Dn=|a1,a2,...,an|
则 |a1+a2,a2+a3,...,an+a1|=|(a1,a2,...,an)K|
其中K=
1 0 0 ...0 1
1 1 0 ...0 0
0 1 1 ...0 0
......
0 0 0 ...1 0
0 0 0 ...1 1
由 |K|=1+(-1)^(n-1)
得 |a1+a2,a2+a3,...,an+a1|=|a1,a2,...,an||K|=[1+(-1)^(n-1)]a.