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,恰为等比数列其中k1=1,k2=6,k3=26﹙1﹚求数列﹛kn﹜的通项公式 ﹙2﹚求数列的﹛kn﹜的前n项和

a1/a6=a6/a26即a1/(a1+5d)=(a1+5d)/(a1+25d),解得25倍d的平方=15（a1)d,
d=0.6a1,an=a1+(n-1)d=(0.6n+0.4)a1.等比数列{ak}的公比q=a6/a1=4=akn/ak(n-1)=[a1+(kn-1)d]/{a1+[k(n-1)-1]d},解得kn*d-4[k(n-1)]*d=3(a1-d)=1.2a1
kn=4k(n-1)+2=4[4k(n-2)+2]+2=…=k1*4^(n-1)+4n+2=4^(n-1)+4n+2
故kn=4^(n-1)+4n+2，
Skn=1*（4^n-1)/3+4n(n+1)/2+2n=（4^n-)/3+n(n+1)+2n=（4^n-1)/3+n^2+3n

(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

(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