設(shè)公比為q.
a2+a3+a4=28 2(a3+2)+a3=28
3a3=24 a3=8
2(a3+2)=a2+a4
2a3+4=a3/q +a3q
a3=8代入,整理,得
2q²-5q+2=0
(2q-1)(q-2)=0
q=1/2(數(shù)列是遞增數(shù)列,舍去)或q=2
an=a1q^(n-1)=a3q^(n-3)=8×2^(n-3)=2ⁿ
數(shù)列{an}的通項公式為an=2ⁿ
bn=anlog(1/2)(an)=2ⁿlog(1/2)(2ⁿ)=-n×2ⁿ
Sn=b1+b2+...+bn=-(1×2+2×2²+3×2³+...+n×2ⁿ)
2Sn=-[1×2²+2×2³+...+(n-1)×2ⁿ+n×2^(n+1)]
Sn-2Sn=-Sn=-[2+2²+...+2ⁿ-n×2^(n+1)]
Sn=2×(2ⁿ-1)/(2-1) -n×2^(n+1)
=(1-n)×2^(n+1)-2
Sn+n×2^(n+1)>30
(1-n)×2^(n+1)-2+n×2^(n+1)>30
2^(n+1)>32
2^(n+1)>2^5
n+1>5
n>4,n為正整數(shù),n≥5,n的最小值是5.