已知数列{an}为等差数列,公差d≠0,{an}的部分项组成下列数列：ak1,ak2,…,akn,恰为等比数列

(1)令an=a1+(n-1)d式1,ak3/ak2=ak2/ak1=q式2
由题,有ak1=a1,ak2=a6,ak3=a26,
由式2,有a1(a1+25d)=(a1+5d)^2,解得a1=5/3d;
带入式1,a1=5/3d,a6=20/3d,a26=80/3d,由式2得q=4;
故有akn=5/3d+(kn-1)d=(5/3d)*q^(n-1),即5/3+kn-1=5/3*4^(n-1);
解之得,kn=5/3*4^(n-1)-2/3.
(2)Sn=5/3*[1*(4^n-1)]/(4-1) - 2/3n=5/9 * (4^n-1)-2/3n