问题
解答题
已知α为锐角,且tanα=
(1)求函数f(x)的表达式; (2)求证:an+1>an; (3)求证:1<
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答案
(1)tan2α=
=2tanα 1-tan2α
=1,2(
-1)2 1-(
-1)22
又∵α为锐角,所以2α=
,π 4
∴sin(2α+
)=1,π 4
则f(x)=x2+x;
(2)∵an+1=f(an)=an2+an,
∴an+1-an=an2>0,
∴an+1>an;
(3)∵
=1 an+1
=1
+ana 2n
=1 an(1+an)
-1 an
,且a1=1 1+an
,1 2
∴
=1 1+an
-1 an
,1 an+1
则
+1 1+a1
+…+1 1+a2
=1 1+an
-1 a1
+1 a2
-1 a2
+…+1 a3
-1 an 1 an+1
=
-1 a1
=2-1 an+1
,1 an+1
∵a2=(
)2+1 2
=1 2
,a3=(3 4
)2+3 4
>1,3 4
又n≥2时,∴an+1>an,
∴an+1≥a3>1,
∴1<2-
<2,1 an+1
∴1<
+1 1+a1
+…+1 1+a2
<2.1 1+an