已知{an}为等差数列,公差d≠0,{an}的部分项ak1,ak2,…,akn恰为等比数列,若k1=1,k2=5,k3=13,

(1):
a5=a1*q=a1+4d a13=a5*q=a1*q*q=a1+12d
(a1+12d)/(a1+4d)=(a1+4d)/a1
1+8d/(a1+4d)=1+4d/a1
8d/(a1+4d)=4d/a1
2a1=a1+4d
a1=4d
所以 a1=4d a5=8d a13=16d kn=2^(n+1)-3
(2)
k1+2k2+3k3+…+nkn=1+2*(2^3-3)+3*(2^4-3)+……+n*[2^(n+1)-3]
=1+[2*2^3+3*2^4+……+n*2^(n+1)]-3(2+3+4……+n)