Question

已知数列{an}是等比数列,前n项和为Sn,a1+2a2=0,S4-S2=1/8？

Answer 1

设公比为q,a1+2a2=0,S4-S2=1/8∴a1+2a1q=0a1q²+a1a³=1/8∴q=-1/2,a1=1∴an=(-1/2)^(n-1)Sn=[1-(-1/2)ⁿ]/(1+1/2)=2/3[1-(-1/2)ⁿ]an*Sn=2/3[(-1/2)^(n-1)-(-1/2)^(2n-1)]{(-1/2)^(2n-1)}是等比...,1,a2=a1*q,a1=-2a2,所以q=-1/2
S4-S2=1/8,所以a3+a4=1/8
a1*(q2+q3)=1/8,a1=1
an*Sn={1-(-2)n}/(1+1/2)=2/3{1/(-2)n-1},2,a1+2a1*q = 0
如果a1=0的话，整个数列全0，S4-S2=0不满足条件，所以q = -1/2
S4-S2= a4+ a3 = a3*(1+q) = 1/8 ==>a3=1/4, a1=1, an=(-1/2) ^(n-1)
对于an*Sn这类问题，需要用乘以公比，然后错位相减，具体过程省略。,1,已知数列{an}是等比数列,前n项和为Sn,a1+2a2=0,S4-S2=1/8
求数列（an*sn）的前n项和