已知数列{an}前n项和为sn,且sn=2n^2+n数列{bn}满足an=4log2（bn）+3,n∈N*1 求an,bn2 求数列{an·bn}的前N项和3设Cn=1/[(an*an+1](n+1为下标）,数列{Cn}前n项和为Tn,且Tn

1.
n=1时,a1=S1=2×1²+1=3
n≥2时,an=Sn-S(n-1)=2n²+n-[2(n-1)²+(n-1)]=4n-1
n=1时,a1=4×1-1=3,同样满足通项公式
数列{an}的通项公式为an=4n-1
an=4log2(bn) +3
log2(bn)=(an -3)/4=(4n-1-3)/4=n-1
bn=2^(n-1)
数列{bn}的通项公式为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)
则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)]
随n增大,4(4n+3)单调递增,1/[4(4n+3)]单调递减,1/12 -1/[4(4n+3)]单调递增
n->+∞,Tn1/12
要不等式Tn