题目内容

等差数列{an}的公差d≠0,它的部分项依次组成的数列,,,…,,…,成等比数列,其中k1=1,k2=5,k3=17.

(1)求等比数列,,,…,,…的公比q.

(2)kn,并求k1+k2+…+kn.

 

答案:
解析:

(1)∵{an}成等差数列.

=a1

=a5=a1+4d

=a17=a1+16d

又{}成等比数列.

a52=a1a17即(a1+4d)2=a1(a1+16d)

整理得:16d2=8a1d,又d≠0.

a1=2d, =a1=2d,

=a5=6d

=3

∴{}数列的公比为3.

(2)由(1)得an=a1+(n-1)d=(n+1)d

=(kn+1)d

=a1qn1=2d·3n1

kn=2·3n1-1

k1+k2+…+kn

=(2-1)+(2·3-1)+…+(2·3n1-1)

=2(1+3+…+3n1)-n

=3nn-1

 


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