x2-4x+4=0，
解得兩個相等的實根x1=x2=2，…（3分）
所以設通項an=（c1+c2n）?2n，
由a1=1，a2=2可得：
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所以an=2n-1，n∈N*…（6分）
（2）由an+2=5an+1-6an可知特征方程為：
x2-5x+6=0，x1=2，x2=3…（8分）
所以?an=c1?2n+c2?3n，
由
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得到c1=c2=1，
所以?an=2n+3n，…（9分）
因為{an+1-λan}是等比數列，
所以有（a2-λa1）?（a4-λa3）=（a3-λa2）2λ=2或λ=3…（10分）
當λ=2時，
| an+1?2an | |
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