等差数列{an},公差d≠0,从an中取出部分项组成等比数列.k1=1,k2=2,k3=8

a1*a8=a2²
a8=a1+7d,a5=a1+d
即a1*( a1+7d)=( a1+d) ²
解得5a1=d
故an=a1+(n-1)d=(5n-4)a1,
显然a8=36a1,a2=6a1
q=a2/a1=6,
所以akn=a1*q^(n-1)=a1*6^(n-1)
同时akn=a1+(kn-1)d=(5kn-4)a1,
故(5kn-4)a1 =a1*6^(n-1) ,
5kn-4=6^(n-1) ,
kn=1/5*6^(n-1)+4/5.
因此k1+k2+.+kn =1/5*（1+6+6^2……+6^(n-1)）+4 n /5
=(6^n -1)/25+4 n /5
=(6^n -1+20n)/25.