1.
n=1時,a1=S1=2×1²+1=3
n≥2時,an=Sn-S(n-1)=2n²+n-[2(n-1)²+(n-1)]=4n-1
n=1時,a1=4×1-1=3,同樣滿足通項公式
數(shù)列{an}的通項公式為an=4n-1
an=4log2(bn) +3
log2(bn)=(an -3)/4=(4n-1-3)/4=n-1
bn=2^(n-1)
數(shù)列{bn}的通項公式為bn=2^(n-1)
2.
an·bn=(4n-1)·2^(n-1)=n·2^(n+1) -2^(n-1)
Kn=a1·b1+a2·b2+...+an·bn
=1·2²+2·2³+3·2⁴+...+n·2^(n+1) -[1+2+...+2^(n-1)]
令Cn=1·2²+2·2³+3·2⁴+...+n·2^(n+1)
則2Cn=1·2³+2·2⁴+...+(n-1)·2^(n+1)+n·2^(n+2)
Cn-2Cn=-Cn=2²+2³+...+2^(n+1) -n·2^(n+2)
=4·(2ⁿ-1)/(2-1) -n·2^(n+2)
=(1-n)·2^(n+2) -4
Cn=(n-1)·2^(n+2) +4
Kn=Cn -[1+2+...+2^(n-1)]
=(n-1)·2^(n+2) +4 -1·(2ⁿ-1)/(2-1)
=(4n-5)·2ⁿ +5
3.
cn=1/[an·a(n+1)]=1/[(4n-1)(4(n+1)-1)]=(1/4)[1/(4n-1)-1/(4(n+1)-1)]
Tn=c1+c2+...+cn
=(1/4)[1/(4×1-1)-1/(4×2-1)+1/(4×2-1)-1/(4×3-1)+...+1/(4n-1)-1/(4(n+1)-1)]
=(1/4)[1/3 -1/(4n+3)]
=1/12 -1/[4(4n+3)]
隨n增大,4(4n+3)單調(diào)遞增,1/[4(4n+3)]單調(diào)遞減,1/12 -1/[4(4n+3)]單調(diào)遞增
n->+∞,Tn1/12
要不等式Tn