设等差数列an的前n项和为Sn,等比数列bn的前n项和为Tn,已知数列bn的公比为q(q>0),a1=b1=1,S5=4
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设等差数列an的前n项和为Sn,等比数列bn的前n项和为Tn,已知数列bn的公比为q(q>0),a1=b1=1,S5=45,T3=
a3-b2.
求q/(a1a2)+q/(a2a3)+…+q/(an+a(n+1))
a3-b2.
求q/(a1a2)+q/(a2a3)+…+q/(an+a(n+1))
设{an}公差为d
S5=5a1+10d=5(a1+2d)=5a3=45
a3=9
a3-a1=2d=9-1=8
d=4
an=a1+(n-1)d=1+4(n-1)=4n-3
T3=b1+b2+b3=b1(1+q+q²)=1×(1+q+q²)=1+q+q²=a3-b2=9-b1q=9-1×q=9-q
q²+2q-8=0
(q+4)(q-2)=0
q=-4(与已知q>0矛盾,舍去)或q=2
q/[ana(n+1)]=2/[(4n-3)(4n+1)]=(1/2)[1/(4n-3) -1/[4(n+1)-3]]
q/(a1a2)+q/(a2a3)+...+q/[ana(n+1)]
=(1/2)[1/(4×1-3)-1/(4×2-3)+1/(4×2-3)-1/(4×3-3)+...+1/(4n-3)-1/[4(n+1)-3]]
=(1/2)[1-1/(4n+1)]
=2n/(4n+1)
S5=5a1+10d=5(a1+2d)=5a3=45
a3=9
a3-a1=2d=9-1=8
d=4
an=a1+(n-1)d=1+4(n-1)=4n-3
T3=b1+b2+b3=b1(1+q+q²)=1×(1+q+q²)=1+q+q²=a3-b2=9-b1q=9-1×q=9-q
q²+2q-8=0
(q+4)(q-2)=0
q=-4(与已知q>0矛盾,舍去)或q=2
q/[ana(n+1)]=2/[(4n-3)(4n+1)]=(1/2)[1/(4n-3) -1/[4(n+1)-3]]
q/(a1a2)+q/(a2a3)+...+q/[ana(n+1)]
=(1/2)[1/(4×1-3)-1/(4×2-3)+1/(4×2-3)-1/(4×3-3)+...+1/(4n-3)-1/[4(n+1)-3]]
=(1/2)[1-1/(4n+1)]
=2n/(4n+1)
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