求证下列行列式（1）a23...n1a+13...n12a+2...n=[a+(n-1)(n-2)](a-1)^(n-1)::::123...a+n-1（2）a+bab0...001a+bab...00n+101a+b...00=∑a^i*b^(n+1-i):::::i=0000a+bab000...1a+b

Question

求证下列行列式
（1）a 2 3 ...n
1 a+1 3 ...n
1 2 a+2 ...n = [a+(n-1)(n-2)](a-1)^(n-1)
::::
1 2 3 ...a+n-1
（2）a+b ab 0 ...0 0
1 a+b ab ...0 0 n+1
0 1 a+b ...0 0 = ∑a^i*b^(n+1-i)
:::::i=0
0 0 0 a+b ab
0 0 0 ...1 a+b

Answer 1

结论都错了.你自己推导的结果吧.
(1)
[a+(n-1)(n+2)/2](a-1)^(n-1)
(2)
(b^(n+1)-a^(n+1))/(b-a)