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))
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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))](/uploads/image/z/4315981-13-1.jpg?t=%E8%AE%BE%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97bn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97bn%E7%9A%84%E5%85%AC%E6%AF%94%E4%B8%BAq%EF%BC%88q%EF%BC%9E0%EF%BC%89%2Ca1%3Db1%3D1%2CS5%3D45%2CT3%3Da3-b2.%E6%B1%82q%2F%EF%BC%88a1a2%EF%BC%89%2Bq%2F%EF%BC%88a2a3%EF%BC%89%2B%E2%80%A6%2Bq%2F%EF%BC%88an%2Ba%28n%2B1%29%EF%BC%89)
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))
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))
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))
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=5*(a1+a5)/2=45 得 a5=17 所以d=(17-1)/4=4
则 an=1+(n-1)*4=4n-3
T3=1+q+q^2=9-q 因为q>0, 得q=2
于是 q/(a1a2)+q/(a2a3)+…+q/(an+a(n+1))
=2*(1/(1*5)+1/(5*9)+....1/(4(n-1)*4n))
=1/2*(1-1/5+1/5-1/9+.....+1/4(n-1)-1/4n)
=1/2*(1-1/4n)