设曲线Cn:f(x)=xn+1(n∈N*)在点P(-12,f(-12))处的切线与y轴交于点Qn(0,yn).(Ⅰ)求数列{yn}的通项公式;(Ⅱ)设数列{yn}的前n项和为Sn,猜测Sn的最大值并证

题目简介

设曲线Cn:f(x)=xn+1(n∈N*)在点P(-12,f(-12))处的切线与y轴交于点Qn(0,yn).(Ⅰ)求数列{yn}的通项公式;(Ⅱ)设数列{yn}的前n项和为Sn,猜测Sn的最大值并证

题目详情

设曲线Cn:f(x)=xn+1(n∈N*)在点P(-
1
2
,f(-
1
2
))
处的切线与y轴交于点Qn(0,yn).
(Ⅰ)求数列{yn}的通项公式;
(Ⅱ)设数列{yn}的前n项和为Sn,猜测Sn的最大值并证明你的结论.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)∵f′(x)=(n+1)xn(n∈N*),(1分)
∴点P处的切线斜率kn=(n+1)(-class="stub"1
2
)n
,(2分)
∴切线方程为:y-(-class="stub"1
2
)n+1=(n+1)(-class="stub"1
2
)n(x+class="stub"1
2
)
,(3分)
令x=0得:yn=(-class="stub"1
2
)n+1+class="stub"n+1
2
•(-class="stub"1
2
)n

故数列{yn}的通项公式为:yn=class="stub"n
2
•(-class="stub"1
2
)n
.(4分)
(Ⅱ)Sn=class="stub"1
2
•(-class="stub"1
2
)+class="stub"2
2
•(-class="stub"1
2
)2+class="stub"3
2
•(-class="stub"1
2
)3++class="stub"n
2
•(-class="stub"1
2
)n

两边同乘-class="stub"1
2
得:-class="stub"1
2
Sn=class="stub"1
2
•(-class="stub"1
2
)2+class="stub"2
2
•(-class="stub"1
2
)3+class="stub"3
2
•(-class="stub"1
2
)4++class="stub"n
2
•(-class="stub"1
2
)n+1

∴得:class="stub"3
2
sn=class="stub"1
2
•(-class="stub"1
2
)+class="stub"1
2
•(-class="stub"1
2
)2+class="stub"1
2
•(-class="stub"1
2
)3++class="stub"1
2
•(-class="stub"1
2
)n-class="stub"n
2
•(-class="stub"1
2
)n+1
(6分)
3Sn=(-class="stub"1
2
)+(-class="stub"1
2
)2+(-class="stub"1
2
)3++(-class="stub"1
2
)n-n•(-class="stub"1
2
)n+1

=
-class="stub"1
2
-(-class="stub"1
2
)
n+1
1+class="stub"1
2
-n•(-class="stub"1
2
)n+1

=-
1-(-class="stub"1
2
)
n
3
-n•(-class="stub"1
2
)n+1

Sn=class="stub"1
9
[class="stub"2+3n
2
(-class="stub"1
2
)
n
-1]
(8分)
其中S1=y1=-class="stub"1
4
,S2=y1+y2=0,S3=-class="stub"3
16
S4=-class="stub"1
16

猜测Sn的最大值为S2=0.证明如下:(10分)
(i)当n为奇数时,Sn=-class="stub"1
9
[class="stub"2+3n
2
(class="stub"1
2
)
n
+1]<0
;(11分)
(ii)当n为偶数时,Sn=class="stub"1
9
•(class="stub"2+3n
2n+1
-1)

h(n)=class="stub"2+3n
2n+1
,则h(n+2)=class="stub"8+3n
2n+3

h(n+2)-h(n)=class="stub"8+3n
2n+3
-class="stub"2+3n
2n+1
=-class="stub"9n
2n+3
<0

∴h(n+2)<h(n).(13分)
h(n)=class="stub"2+3n
2n+1
的最大值为h(2)=1,即Sn的最大值为S2=0.(14分)
解法2(Ⅱ)任意k∈N*,都有y2k-1<0,y2k>0;
所以Sn的最大值就是S2k的最大值.
ak=y2k-1+y2k=class="stub"1-k
4k
,显然a1=0,k>1,ak<0,
所以S2k=a1+a2++ak的最大值是S2=a1=0.

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