c2+c1、c3+c2、...、c(n+1)+cn,行列式成《下三角》,a(n+1)(n+1)=(n+1)b
D(n+1)=|a1 0 0 ...0|
0 a2 0 ...0
0 0 a3 ...0
.
b 2b 3b ...(n+1)b
=(n+1)b*∏ai (i=1 to n)
c2+c1、c3+c2、...、c(n+1)+cn,行列式成《下三角》,a(n+1)(n+1)=(n+1)b
D(n+1)=|a1 0 0 ...0|
0 a2 0 ...0
0 0 a3 ...0
.
b 2b 3b ...(n+1)b
=(n+1)b*∏ai (i=1 to n)