高考数学一轮复习课件--求数列的通项公式

n≤4 时, an≤0; 当 n≥5 时, an>0;Р当 n≥5 时, Tn=-S4+Sn-S4=Sn-2S4Р故 an=Р2n-8, n≥2.Р-14, n=1,Р=n2-7n-8-2(-20)Р∴当 n≤4 时, Tn=-Sn=-n2+7n+8,Р=n2-7n+32.Р故 Tn=Рn2-7n+32, n≥5.Р-n2+7n+8, n≤4,Р例4 求下列数列的通项公式:? (1)2,22,222,2222,· · · (逐项依次多数字2)? (2)0.23,0.2323,0.232323,· · · (逐项依次多数字23).Р(2)解法同上,此小题留给同学们完成,其答案?为:Р4、化归(构造)法Р通过恰当的恒等变形, 如配方、因式分解、取倒数等, 转化为等比数列或等差数列.Р(1)若 an+1=pan+q, 则:Рan+1-=p(an-).Р(3)若 an+1=pan+q(n), 则:Р(2)若 an+1= , 则:РpanРr+qanРan+1Р1РanР1Р= · + .РpРrРpРqРan+1Рpn+1РanРpnР= + .Рq(n)Рpn+1Р例5.已知数列{an} 中, a1=1, an+1= an+1(nN*), 求 an.Р1Р2Р解法一∵an+1= an+1(nN*),Р1Р2Р∴an= an-1+1, an-1= an-2+1.Р1Р2Р1Р2Р两式相减得: an-an-1= (an-1-an-2)Р1Р2Р∴{an-an-1} 是以 a2-a1= 为首项, 公比为的等比数列.Р1Р2Р1Р2Р∴an-an-1= ( )n-2=( )n-1.Р1Р2Р1Р2Р1Р2Р∴an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)Р=1+ +( )2+…+( )n-1Р1Р2Р1Р2Р1Р2Р=2-21-n.Р即 an=2-21-n.