题目内容
在直角坐标平面xoy上 的一列点A1(1,a1),A2(2,a2),…,An(n,an),…,简记为{An}.若由bn=
•
构成的数列{bn}满足bn+1>bn(其中
是y轴正方向同向的单位向量),则称{An}为T点列.
(1)判断A1(1,1),A2(2,
),A3(3,
)…,An(n,
),…是否为T点列;
(2)若{an}是等差数列,判断点列A1(1,a1),A2(2,a2),…,An(n,an),…是否为T点列,并说明理由;
若{an}是等比数列,判断点列A1(1,a1),A2(2,a2),…,An(n,an),…是否为T点列,并说明理由;
(3)若{An}为T点列,且点A2在点A1的右上方,任取其中连续三点AK,AK+1,AK+2,判断△AKAK+1AK+2的形状(锐角三角形,直角三角形,钝角三角形),并说明理由.
AnAn+1 |
j |
j |
(1)判断A1(1,1),A2(2,
1 |
2 |
1 |
3 |
1 |
n |
(2)若{an}是等差数列,判断点列A1(1,a1),A2(2,a2),…,An(n,an),…是否为T点列,并说明理由;
若{an}是等比数列,判断点列A1(1,a1),A2(2,a2),…,An(n,an),…是否为T点列,并说明理由;
(3)若{An}为T点列,且点A2在点A1的右上方,任取其中连续三点AK,AK+1,AK+2,判断△AKAK+1AK+2的形状(锐角三角形,直角三角形,钝角三角形),并说明理由.
分析:(1)根据所给的n个点的坐标,求得bn=
•
=-
,再验证数列{bn}是否满足满足bn+1>bn,即可得到结论.
(2)设等差数列的公差为d,根据An(n,an),An+1(n+1,an+1),求得bn=
•
=d即可判断;设等比数列的公比为q,根据An(n,an),An+1(n+1,an+1),可得bn=
•
=an+1-an,从而可得结论;
(3)用所给的三个点构造三个向量,写出三个向量的坐标,问题转化为向量夹角的大小问题,判断出两个向量的数量积小于零,得到两个向量所成的角是钝角,得到结果.
AnAn+1 |
j |
1 |
n(n+1) |
(2)设等差数列的公差为d,根据An(n,an),An+1(n+1,an+1),求得bn=
AnAn+1 |
j |
AnAn+1 |
j |
(3)用所给的三个点构造三个向量,写出三个向量的坐标,问题转化为向量夹角的大小问题,判断出两个向量的数量积小于零,得到两个向量所成的角是钝角,得到结果.
解答:解:(1)∵An(n,
),An+1(n+1,
),
∴
=(1,-
),
又∵
=(0,1),∴bn=
•
=-
,
∴bn+1=-
,∴bn+1-bn=
∴bn+1>bn,∴{An}是T点列;
(2)设等差数列的公差为d
∵An(n,an),An+1(n+1,an+1),
∴
=(1,an+1-an)=(1,d),
又∵
=(0,1),∴bn=
•
=d,∴{An}不是T点列;
设等比数列的公比为q,
∵An(n,an),An+1(n+1,an+1),
∴
=(1,an+1-an),
又∵
=(0,1),∴bn=
•
=an+1-an,
∴bn+1=an+2-an+1=qbn,
∴q>1时,bn+1>bn,∴{{An}是T点列;q≤1时,bn+1≤bn,∴{{An}不是T点列;
(3)在△AKAK+1AK+2中,Ak+1Ak=(-1,ak-ak+1),Ak+1Ak+2=(1,ak+2-ak+1),
Ak+1Ak•Ak+1Ak+2=-1+(ak+2-ak+1)(ak-ak+1)
∵点A2在点A1的右上方,∴b1=a2-a1>0,
∵{An}为T点列,
∴bn≥b1>0,
∴(ak+2-ak+1)(ak-ak+1)=-bk+1bk<0,则 Ak+1Ak•Ak+1Ak+2<0
∴∠AKAK+1AK+2为钝角,
∴△AKAK+1AK+2为钝角三角形.
1 |
n |
1 |
n+1 |
∴
AnAn+1 |
1 |
n(n+1) |
又∵
j |
AnAn+1 |
j |
1 |
n(n+1) |
∴bn+1=-
1 |
(n+1)(n+2) |
2 |
n(n+1)(n+2) |
∴bn+1>bn,∴{An}是T点列;
(2)设等差数列的公差为d
∵An(n,an),An+1(n+1,an+1),
∴
AnAn+1 |
又∵
j |
AnAn+1 |
j |
设等比数列的公比为q,
∵An(n,an),An+1(n+1,an+1),
∴
AnAn+1 |
又∵
j |
AnAn+1 |
j |
∴bn+1=an+2-an+1=qbn,
∴q>1时,bn+1>bn,∴{{An}是T点列;q≤1时,bn+1≤bn,∴{{An}不是T点列;
(3)在△AKAK+1AK+2中,Ak+1Ak=(-1,ak-ak+1),Ak+1Ak+2=(1,ak+2-ak+1),
Ak+1Ak•Ak+1Ak+2=-1+(ak+2-ak+1)(ak-ak+1)
∵点A2在点A1的右上方,∴b1=a2-a1>0,
∵{An}为T点列,
∴bn≥b1>0,
∴(ak+2-ak+1)(ak-ak+1)=-bk+1bk<0,则 Ak+1Ak•Ak+1Ak+2<0
∴∠AKAK+1AK+2为钝角,
∴△AKAK+1AK+2为钝角三角形.
点评:本题考查数列和不等式的综合运用,解题时要认真审题,注意T点列的理解和合理地进行等价转化.

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