等差数列{an}的公差为-2,且a1,a3,a4成等比数列.设bn=2/n(12-an)(n∈N*),求数列{bn}的前n项和Sn?
问题描述：
等差数列{an}的公差为-2,且a1,a3,a4成等比数列.设bn=2/n(12-an)(n∈N*),求数列{bn}的前n项和Sn?
答
a1,a3,a4成等比数列 a3=a1+(3-1)*(-2)=a1-4 ; a4=a1-6a3*a3=a1*a4(a1-4)(a1-4)=a1(a1-6)a1*a1-8a1+16=a1*a1-6a12a1=16a1=8则an=10-2n bn=2/n(12-an)=2/n(12-(10-2n))=2/n(2+2n)=1/n(n+1)=1/n-1/(n+1)Sn=b1+b2+...+bn...打错了，是bn=1/n(12-an)(n∈N*),bn=1/n(12-an)=1/n(12-(10-2n))=1/n(2+2n)=1/2n(n+1)=(1/2)[1/n-1/(n+1)] Sn=b1+b2+...+bn =(1/2)(1/1-1/2)+(1/2)(1/2-1/3)+...+(1/2)[1/n-1/(n+1)] =(1/2)[1-1/2+1/2-1/3+...+1/n-1/(n+1)] =(1/2)[1-1/(n+1)] =n/2(n+1)