问题
解答题
已知曲线C:xy=1,过C上一点An(xn,yn)作一斜率为kn=-
(1)求xn与xn+1的关系式; (2)求证:{
(3)求证:(-1)x1+(-1)2x2+(-1)3x3+…+(-1)nxn<1(n∈N,n≥1). |
答案
(1)过C:y=
上一点An(xn,yn)作斜率为kn的直线交C于另一点An+1,1 x
则kn=
=yn+1-yn xn+1-xn
=-
-1 xn+1 1 xn xn+1-xn
=-1 xn+1•xn
,1 xn+2
于是有:xnxn+1=xn+2
即:xn+1=1+
.2 xn
(2)记an=
+1 xn-2
,1 3
则an+1=
+1 xn+1-2
=1 3
+1
-2xn+2 xn
=-2(1 3
+1 xn-2
)=-2an,1 3
因为x1=
, 而a1=11 7
+1 x1-2
=-2≠0,1 3
因此数列{
+1 xn-2
}是等比数列.1 3
(3)由(2)知:an=(-2)n , 则xn=2+
,(-1)nxn=(-1)n•2+1 (-2)n- 1 3
.1 2n-(-1)n• 1 3
①当n为偶数时有:(-1)n-1xn-1+(-1)nxn=
=
+1 2n-1+ 1 3
=1 2n- 1 3
<2n-1+2n (2n-1+
)(2n-1 3
)1 3
<2n-1+2n 2n-1•2n
+1 2n-1
,1 2n
于是在n为偶数时有:(-1)x1+(-1)2x2++(-1)nxn<
+1 2
+1 22
+1 23
++1 24
<1.1 2n
1在n为奇数时,前n-1项为偶数项,
于是有:(-1)x1+(-1)2x2++(-1)n-1xn-1+(-1)nxn<1+(-1)nxn=1-xn=1-(2+
)=-1+1 (-2)n- 1 3
<1.1 2n+ 1 3
综合①②可知原不等式得证.