已知数列满足递推式an=2a(n-1)+1(n>=2),其中a3=7.求通项公式;已知数列bn满足bn=n/(2n+1),求bn的前n项和S

题目

已知数列满足递推式an=2a(n-1)+1(n>=2),其中a3=7.求通项公式;已知数列bn满足bn=n/(2n+1),求bn的前n项和S

答案

1.
a3=2a2+1 a2=(a3-1)/2=(7-1)/2=3
a2=2a1+1 a1=(a2-1)/2=(3-1)/2=1
n≥2时,
an=2a(n-1)+1
an +1=2a(n-1)+2=2[a(n-1)+1]
(an +1)/[a(n-1)+1]=2,为定值.
a1 +1=1+1=2
数列{an}是以2为首项,2为公比的等比数列.
an +1=2×2^(n-1)=2ⁿ
an=2ⁿ-1
n=3时,a3=2³-1=8-1=7,同样满足
数列{an}的通项公式为an=2ⁿ-1.
2.
bn=n/(an+1)=n/(2ⁿ-1+1)=n/2ⁿ
Sn=b1+b2+...+bn=1/2+ 2/2²+3/2³+...+n/2ⁿ
(1/2)Sn=1/2²+2/2³+...+(n-1)/2ⁿ+n/2^(n+1)
Sn-(1/2)Sn=(1/2)Sn
=1/2+1/2²+...+1/2ⁿ -n/2^(n+1)
=(1/2)(1-1/2ⁿ)/(1-1/2) -n/2^(n+1)
=1- 1/2ⁿ-n/2^(n+1)
Sn=2 -(n+2)/2ⁿ