题目内容
各项均为正数的等比数列{an}满足a1a7=4,a6=8,函数f(x)=a1x+a2x2+a3x3+…+a10x10,则f(
)=______.
| 1 |
| 2 |
∵正数的等比数列{an}满足a1a7=4,∴
=4,可得a4=2,
∵a6=8,
∴
=q2,可得q2=4,可得q=2,∴a1×q3=2,得a1=
,
∴an=a1×qn=
×2n-1=2n-3,
∴f(x)=a1x+a2x2+a3x3+…+a10x10,
∴f(
)=a1
+a2
2+a3(
)3+…+a10(
)10=
+
+…+
=10×
=
=
,
故答案为
;
| a | 24 |
∵a6=8,
∴
| a6 |
| a4 |
| 1 |
| 4 |
∴an=a1×qn=
| 1 |
| 4 |
∴f(x)=a1x+a2x2+a3x3+…+a10x10,
∴f(
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 23 |
| 1 |
| 23 |
| 1 |
| 23 |
| 1 |
| 23 |
| 10 |
| 8 |
| 5 |
| 4 |
故答案为
| 5 |
| 4 |
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