k |
x2 |
k |
x3 |
k |
x4 |
k |
x2010 |
k2009 |
x2010 |
又因为x1=1,所以原式=
k2009 |
x2010 |
k2009 |
2010 |
又因为T1=
1 |
2 |
1 |
2 |
1 |
2 |
k |
x2 |
于是T1•T2•…•T2009=
k2009 |
2010 |
1 |
2010 |
故答案为:
1 |
2010 |
k |
x |
1 |
2 |
k |
x2 |
k |
x3 |
k |
x4 |
k |
x2010 |
k2009 |
x2010 |
k2009 |
x2010 |
k2009 |
2010 |
1 |
2 |
1 |
2 |
1 |
2 |
k |
x2 |
k2009 |
2010 |
1 |
2010 |
1 |
2010 |