1,已知a+1/a=2,求a^2+!/a^2,猜想a^n+1/a^n(n是任意一个自然数）等于几?并证明你的结论.2,计算：1/a-1+1/(a-1)(a-2)+1/(a-2)(a-3)+...+1/(a-2004)(a-2005)

1.a+1/a=2,a^2-2a+1=0,(a-1)^2=0,a=1,a^2+1/a^2=1+1=2,a^n+1/a^n=2
2.1/(a-1)+1/(a-1)(a-2)+1/(a-2)(a-3)+...+1/(a-2004)(a-2005)
=1/(a-1)+1/(a-2)-1/(a-1)+1/(a-3)-1/(a-4)+...+1/(a-2005)-1/(a-2004)
=1/(a-2005)

1.
a+1/a=2
a^2+1/a^2=(a+1/a)^2-2=4-2=2
a^3+1/a^3=(a+1/a)(a^2+1/a^2)-(a+1/a)=2*2-2=2
猜想a^n+1/a^n=2
证明：（数学归纳法）
设n=k时成立,即a^k+1/a^k=2,a^(k-1)+1/a^(k-1)]=2
a^(k+1)+1/a^(k+1)=[a^k+1/a^k](a+1/a)-[a^(k-1)+1/a^(k-1)]
=2*2-2=2
即 n=k+1亦成立
证明a^n+1/a^n=2 成立
2.
1/a-1+1/(a-1)(a-2)+1/(a-2)(a-3)+...+1/(a-2004)(a-2005)
=1/a-1+[1/(a-2)-1/(a-1)]+[1/(a-3)-1/(a-2)]+...+[1/(a-2005)-1/(a-2004)]
=1/(a-2005)