等比数列的前n项和公式递推公式求通项 高二上学期数学人教A版（2019）选择性必修第二册

递推公式求通项4.3.2等比数列的前n项和公式特殊值法排除法归纳法求解方法递推公式求通项解答题填空题选择题定义叠加累积辅助数列an+1-an=dan+1-an=f(n)an+1÷an=f(n)可求和可求积an+1=can+f(n)an+1÷an=q等差数列等比数列an+1=can+d，an+1=can+dn+e，an+1=can+d·qn学习目标：求an.例1.已知数列{an}满足a1=1,an+1=2an+3,求an.解法1：待定系数法设an+1+t=2(an+t)an+1=2an+2t-tan+1=2an+tan+1=2an+3由an+1=2an+3得an+1+3=2(an+3)∵a1+3=4≠0∴数列{an+3}为等比数列，首项为4，公比为2.∴an+3=4×2n-1=2n+1∴an=2n+1-3.例1.已知数列{an}满足a1=1,an+1=2an+3,求an.解法2：由an+1=2an+3得an+2=2an+1+3∵a2-a1=a1+3=4≠0∴数列{an+1-an}为等比数列，首项为4，公比为2.∴an+1-an=4×2n-1=2n+1∴an=a1+(a2-a1)+(a3-a2)+···+(an-an-1)=1+22+23+24+···+2n…①…②②-①得an+2-an+1=2(an+1-an)设bn=an+1-anb1=a2-a1=1+3=4≠0则bn+1=2bn∴数列{bn}为等比数列，∴bn=4×2n-1=2n+1首项为4，公比为2.例1.已知数列{an}满足ca1-c+d≠0,an+1=can+d,求an.解法1：待定系数法设an+1+t=c(an+t)an+1=can+(c-1)tan+1=can+d由an+1=can+d∴数列为等比数列，首项为，公比为c.例2.已知数列{an}满足a1=1,an+1=2an+3n+1,求an.解法1：待定系数法设an+1+s(n+1)+t=2(an+sn+t)an+1=2an+sn+t-san+1=2an+3n+1由an+1=2an+3n+1∵a1+3+4=8≠0∴{an+3n+4}为等比数列，∴an+3n+4=8×2n-1=2n+2∴an=2n+2-3n-4.s=3,t=4得an+1+3(n+1)+4=2(an+3n+4)首项为8，公比为2.例2.已知数列{an}满足a1=1,an+1=2an+3n+1,求an.解法2：由an+1=2an+3n+1得an+2=2an+1+3(n+1)+1…①…②②-①得an+2-an+1=2(an+1-an)+3设bn=an+1-an则bn+1=2bn+3,即bn+1+3=2(bn+3),b1+3=8.∴数列{bn+3}为等比数列，首项为8，公比为2,则bn+3=8×2n-1=2n+2∴bn=2n+2-3即an+1-an=2n+2-3.∴an=a1+(a2-a1)+(a3-a2)+···+(an-an-1)=1+23+24+···+2n+1-3(n-1)例3.已知数列{an}满足a1=1,an+1=2an+3n,求an.解法1：待定系数法设an+1+t·3n+1=2(an+t·3n)an+1=2an+2t·3n-t·3n+1an+1=2an-t·3nan+1=2an+3n得an+1-3n+1=2(an-3n)∵a1-3=-2≠0∴数列{an-3n}为等比数列，首项为-2，公比为2.∴an-3n=-2×2n-1=-2n∴an=3n-2n.t=-1由an+1=2an+3n例3.已知数列{an}满足a1=1,an+1=2an+3n,求an.解法2：两边同除以3n得由an+1=2an+3n∴数列{bn-3}为等比数列，首项为-2，公比为例3.已知数列{an}满足a1=1,an+1=2an+3n,求an.解法3：两边同除以2n得由an+1=2an+3n∴bn=b1+(b2-b1)+(b3-b2)+···+(bn-bn-1)例4.已知数列{an}满足a1=1,an+1=3an+3n,求an.待定系数法设an+1+t·3n+1=3(an+t·3n)an+1=3an+3t·3n-t·3n+1an+1=3anan+1=3an+3nt不存在由an+1=3an+3n解：两边同除以3n得∴数列{bn}为等差数列，首项为1，公差为1.∴bn=n.∴an=n·3n-1.则bn+1=bn+1,即