应该消掉a1,.,an,留下2^2,.,2^n等
Tn=2an＋2²a(n-1)＋2∧3a(n-2)＋…＋2∧na1①
2Tn=2²an＋2∧3a(n-1)＋2∧4a(n-2)＋…2∧(n＋1)a1②
错位相减法an=3n-1即an-a(n-1）=a(n-1)-a(n-2)=.=a2-a1=3
②-①Tn=-2an＋2²[an-a(n-1)]＋2∧3[a(n-1)-a(n-2)]＋…＋2∧n[a2-a1]+2∧(n＋1)a1
=-2an＋3*2²＋3*2∧3＋…＋3*2∧n+2∧(n＋1)*2
=-2an＋3(2²＋2∧3＋…＋2∧n)+2∧(n＋2)
=-2an+3[-4+2^(n+1)]+2∧(n＋2)
=-2(3n-1)-12+3*2^(n+1)+2*2∧(n＋1)
=5*2^(n+1)-6n+2-12
左边=Tn+12=5*2^(n+1)-6n+2-12+12=5*2^(n+1)-6n+2
右边=-2an+10bn
=-2（3n-1）+10*2^n
=-2（3n-1）+5*2*2^n
=-6n+2+5*2^(n+1)
即证
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