已知递增的正项等比数列{an}中,a5-a1=15,a4-a2=6求an、sn详解,谢谢!

Question

已知递增的正项等比数列{an}中,a5-a1=15,a4-a2=6 求an、sn
详解,谢谢!

Answer 1

设数列公比为q,数列为递增数列,则q>1
a5-a1=15
a1q⁴-a1=15 (1)
a4-a2=6
a1q³-a1q=6 (2)
(1)/(2)
(a1q⁴-a1)/(a1q³-a1q)=15/6
(q⁴-1)/[q(q²-1)]=5/2
(q²+1)(q²-1)/[q(q²-1)]=5/2
(q²+1)/q=5/2
2q²-5q+2=0
(2q-1)(q-2)=0
q=1/2(舍去)或q=2
q=2代入(1)
a1=15/(q⁴-1)=15/(2⁴-1)=15/(16-1)=15/15=1
an=a1q^(n-1)=1×2^(n-1)=2^(n-1)
Sn=a1(qⁿ-1)/(q-1)=1×(2ⁿ-1)/(2-1)=2ⁿ-1