應(yīng)該消掉a1,.,an,留下2^2,.,2^n等
Tn=2an＋2 a(n-1)＋2∧3a(n-2)＋…＋2∧na1①
2Tn=2 an＋2∧3a(n-1)＋2∧4a(n-2)＋…2∧(n＋1)a1②
錯(cuò)位相減法an=3n-1即an-a(n-1）=a(n-1)-a(n-2)=.=a2-a1=3
②-①Tn=-2an＋2 [an-a(n-1)]＋2∧3[a(n-1)-a(n-2)]＋…＋2∧n[a2-a1]+2∧(n＋1)a1
=-2an＋3*2 ＋3*2∧3＋…＋3*2∧n+2∧(n＋1)*2
=-2an＋3(2 ＋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
左邊=Tn+12=5*2^(n+1)-6n+2-12+12=5*2^(n+1)-6n+2
右邊=-2an+10bn
=-2（3n-1）+10*2^n
=-2（3n-1）+5*2*2^n
=-6n+2+5*2^(n+1)
即證