设数列｛an｝满足an+1/an=q,q不等于0数列[｛bn｝满足bn=na1+(n-1)a2+…＋2an-1＋an,求

∵数列｛an｝满足a(n+1)/an=q
∴{an}是等比数列,公比为q
∵bn=na1+(n-1)a2+…＋2an-1＋an
n=1时,b1=a1,
n=2时,b2=2a1+a2
∵ b1＝1.b2=3/2
∴a1=1,2*1+a2=3/2,a2=-1/2
∴q=a2/a1=-1/2
∴an=(-1/2)^(n-1)
bn=n+(n-1)*(-1/2)+(n-2)*1/4+.+(-1/2)^(n-1) (1)
-1/2bn=n(-1/2)+(n-1)*(-1/2)^2+(n-2)*(-1/2)^3+.+2*(-1/2)^(n-1)+(-1/2)^n (2)
(1)-(2):
3/2bn=n-(-1/2)-(-1/2)^2-(-1/2)^3-.-(-1/2)^(n-1)-(-1/2)^n
=n-(-1/2)[1-(-1/2)^(n-1)]/(3/2)-(-1/2)^n
=n+1/3-1/3*(-1/2)^（n-1）+1/2*(-1/2)^(n-1)
=n+1/3+1/6*(-1/2)^(n-1)
bn=(6n+2)/9+1/9*(-1/2)^(n-1)
bn=[6n+2+(-1/2)^(n-1) ]/9

等于1

由bn=na1+(n-1)a2+…＋2an-1＋an，b1＝1.b2=3/2
得a1=1，
2a1+a2=3/2，即a2=-1/2
又a（n+1)/an=q，
有a2/a1=q=-1/2
所以an=(-1/2)^(n-1)
则bn=na1+(n-1)a2+…＋2an-1＋an=nq^0+(n-1)q^1+... ... +q^(n-1)
q*...