Question

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？

Answer 1

a1,a3,a4成等比数列 a3=a1+(3-1)*(-2)=a1-4 ; a4=a1-6
a3*a3=a1*a4
(a1-4)(a1-4)=a1(a1-6)
a1*a1-8a1+16=a1*a1-6a1
2a1=16
a1=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
=1/1-1/2+1/2-1/3+...+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)

追问

打错了，是bn=1/n(12-an)(n∈N*),

追答

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)