题目内容
数列{an}为正项等比数列,若a2=1,且an+an+1=6an-1(n∈N,n≥2),则此数列的前4项和S4=______.
∵{an}是等比数列,
∴an+an+1=6an-1可化为a1qn-1+a1qn=6a1qn-2,
∴q2+q-6=0.
∵数列{an}为正项等比数列,即q>0,
∴q=2.
又a2=a1q=1,∴a1=
,
∴S4=
=
=
.
故答案为:
∴an+an+1=6an-1可化为a1qn-1+a1qn=6a1qn-2,
∴q2+q-6=0.
∵数列{an}为正项等比数列,即q>0,
∴q=2.
又a2=a1q=1,∴a1=
| 1 |
| 2 |
∴S4=
| a1(1-q4) |
| 1-q |
| ||
| 1-2 |
| 15 |
| 2 |
故答案为:
| 15 |
| 2 |
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