Question

求解，在线等，急急

bn是公比为q的等比数列

Answer 1

a1+a2+...+an=Sn = n^2
b1=1, bn=q^(n-1)
(1)
n=1, a1=1
for n>=2
an =Sn -S(n-1) = 2n-1
cn = an +bn
c1=1+1 =2
c2=3+q
c3 = 5+q^2
{cn} 是等差数列
c1+c3 = 2c2
2+5+q^2 = 2(3+q)
q^2-2q+1=0
q=1
bn = 1
cn = an +bn = 2n
(2)
dn = an +n +bn
= 2n-1 +n +1
=3n
Tn =d1+d2+...+dn = (3/2)n(n+1)

追问

有微信吗，还有几题，可以提高悬赏