1.在各项均为正数的等比数列中,若a5·a9=9,求log3a1+log3a2+.+log3a10的值.2.已知数列{an}：an=8/（n+1）（n+3）,求前n项和Sn3.求和：1+3/2^2+4/2^3+.+(n+1)/2^n

1.在各项均为正数的等比数列中,若a5·a9=9,求log3a1+log3a2+.+log3a10的值.2.已知数列{an}：an=8/（n+1）（n+3）,求前n项和Sn3.求和：1+3/2^2+4/2^3+.+(n+1)/2^n
1.在各项均为正数的等比数列中,若a5·a9=9,求log3a1+log3a2+.+log3a10的值.
2.已知数列{an}：an=8/（n+1）（n+3）,求前n项和Sn
3.求和：1+3/2^2+4/2^3+.+(n+1)/2^n

1.在各项均为正数的等比数列中,若a5·a9=9,求log3a1+log3a2+.+log3a10的值.2.已知数列{an}：an=8/（n+1）（n+3）,求前n项和Sn3.求和：1+3/2^2+4/2^3+.+(n+1)/2^n
1.
题目有误,你的这个式子是求不出来的,应该是+...+log3(a13)吧.
a5a9=a7²=9
数列是正项数列,a7>0 a7=3
log3(a1)+log3(a2)+...+log3(a13)
=log3(a1a2...a13)
=log3[(a1a13)(a2a12)...(a6a8)a7]
=log3(a7^13)
=13log3(a7)
=13log3(3)
=13
2.
an=8/[(n+1)(n+3)]=4[1/(n+1)-1/(n+3)]
Sn=a1+a2+...+an
=4[1/2-1/4+1/3-1/5+...+1/(n+1)-1/(n+3)]
=4[(1/2+1/3+...+1/(n+1))-(1/4+1/5+...+1/(n+3))]
=4[1/2+1/3-1/(n+2)-1/(n+3)]
=10/3 -4/(n+2) -4/(n+3)
3.
令Sn=1+3/2²+4/2³+...+(n+1)/2ⁿ=2/2+3/2²+4/2³+...+(n+1)/2ⁿ=
则Sn/2=2/2²+3/2³+...+n/2ⁿ+(n+1)/2^(n+1)
Sn-Sn/2=1+1/2²+...+1/2ⁿ -(n+1)/2^(n+1)
=1×[1-1/2^(n+1)]/(1-1/2) -(n+1)/2^(n+1)
=2-2/2^(n+1) -(n+1)/2^(n+1)
=2-(n+3)/2^(n+1)
Sn=4- (n+3)/2ⁿ
1+3/2²+4/2³+...+(n+1)/2ⁿ=4- (n+3)/2ⁿ

a(n)=aq^(n-1),
9 = a(5)a(6)=a^2q^(9),
log_{3}[a(1)a(2)...a(10)]=log_{3}[a^(10)q^[1+2+...+9]]=log_{3}[a^(10)q^(45)]
=5log_{3}[a^2q^(9)]
=5log_{3}[9]
=5*2
=10
a(n)=8/[(n+1)(n...

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a(n)=aq^(n-1),
9 = a(5)a(6)=a^2q^(9),
log_{3}[a(1)a(2)...a(10)]=log_{3}[a^(10)q^[1+2+...+9]]=log_{3}[a^(10)q^(45)]
=5log_{3}[a^2q^(9)]
=5log_{3}[9]
=5*2
=10
a(n)=8/[(n+1)(n+3)]=8(n+2)/[(n+1)(n+2)(n+3)]=[8(n+3)-8]/[(n+1)(n+2)(n+3)]
=8/[(n+1)(n+2)] - 8/[(n+1)(n+2)(n+3)]
=8/(n+1)-8/(n+2) - 4/[(n+1)(n+2)] + 4/[(n+2)(n+3)]
s(n) = a(1)+a(2)+...+a(n-1)+a(n)
=8/2-8/3 + 8/3-8/4 + ... + 8/n-8/(n+1) + 8/(n+1)-8/(n+2) - 4/[2*3] + 4/[3*4] - 4/[3*4]+4/[5*6] -...-4/[n(n+1)] + 4/[(n+1)(n+2)] - 4/[(n+1)(n+2)] + 4/[(n+2)(n+3)]
=8/2 - 8/(n+2) - 4/[2*3] + 4/[(n+2)(n+3)]
=10/3 + 4[1-2n-6]/[(n+2)(n+3)]
=10/3 - 4(2n+5)/[(n+2)(n+3)]
s(n)=(1+1)/2 + (2+1)/2^2 + ... + [(n-1)+1]/2^(n-1) + (n+1)/2^n,
2s(n)=(1+1) + (2+1)/2 + ... + [(n-1)+1]/2^(n-2) + (n+1)/2^(n-1),
s(n) = 2s(n)-s(n) = (1+1) + 1/2 + ... + 1/2^(n-1) - (n+1)/2^n
= 1 + [1 - 1/2^n]/(1-1/2) - (n+1)/2^n
= 1 + 2 - 1/2^(n-1) - (n+1)/2^n
= 3 - (n+3)/2^n

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