题目

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

提问时间：2020-12-17

答案

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