数列{an}满足a1=1,an=1/2an-1+1(n≥2)求an通项式~

问题描述:

数列{an}满足a1=1,an=1/2an-1+1(n≥2)
求an通项式~

令bn=an-2an-2=1/2an-1-1=1/2(an-1-2)∴(an-2)/(an-1-2)=1/2即bn/bn-1=1/2又有a1=1 b1=-1 ∴{bn}为首项b1=-1,公比是1/2的等比数列bn}通项公式为bn=-(1/2)^(n-1)带入bn=an-2得 an=2-(1/2)^(n-1) (n≥2)a=1时不...