已知数列{an}中的相邻两项a2k-1,a2k是关于x的方程x2-(3k+2k)x+3k•2k=0的两个根,且a2k-1≤a2k(k=1,2,3,…).(I)求a1,a3,a5,a7;(II)求数列{

题目简介

已知数列{an}中的相邻两项a2k-1,a2k是关于x的方程x2-(3k+2k)x+3k•2k=0的两个根,且a2k-1≤a2k(k=1,2,3,…).(I)求a1,a3,a5,a7;(II)求数列{

题目详情

已知数列{an}中的相邻两项a2k-1,a2k是关于x的方程x2-(3k+2k)x+3k•2k=0的两个根,且a2k-1≤a2k(k=1,2,3,…).
(I)求a1,a3,a5,a7
(II)求数列{an}的前2n项和S2n
(Ⅲ)记f(n)=
1
2
(
|sinn|
sinn
+3)
Tn=
(-1)f(2)
a1a2
+
(-1)f(3)
a3a4
+
(-1)f(4)
a5a6
+…+
(-1)f(n+1)
a2n-1a2n
,求证:
1
6
Tn
5
24
(n∈N*)
题型:解答题难度:中档来源:浙江

答案

(I)方程x2-(3k+2k)x+3k•2k=0的两个根为x1=3k,x2=2k,
当k=1时,x1=3,x2=2,所以a1=2;
当k=2时,x1=6,x2=4,所以a3=4;
当k=3时,x1=9,x2=8,所以a5=8时;
当k=4时,x1=12,x2=16,所以a7=12.
(II)S2n=a1+a2++a2n=(3+6++3n)+(2+22++2n)=
3n2+3n
2
+2n+1-2

(III)证明:Tn=class="stub"1
a1a2
+class="stub"1
a3a4
-class="stub"1
a5a6
+…+
(-1)f(n+1)
a2n-1a2n

所以T1=class="stub"1
a1a2
=class="stub"1
6
T2=class="stub"1
a1a2
+class="stub"1
a3a4
=class="stub"5
24
.当n≥3时,Tn=class="stub"1
6
+class="stub"1
a3a4
-class="stub"1
a5a6
+…+
(-1)f(n+1)
a2n-1a2n
≥class="stub"1
6
+class="stub"1
a3a4
-(class="stub"1
a5a6
+…+class="stub"1
a2n-1a2n
)
≥class="stub"1
6
+class="stub"1
6•22
-class="stub"1
6
(class="stub"1
23
+…+class="stub"1
2n
)
=class="stub"1
6
+class="stub"1
6•22
-class="stub"1
24
(1-class="stub"1
2n-3
)> class="stub"1
6

同时,Tn=class="stub"5
24
-class="stub"1
a5a6
-class="stub"1
a7a8
+…+
(-1)f(n+1)
a2n-1a2n
≤class="stub"5
24
-class="stub"1
a5a6
+(class="stub"1
a7a8
+…+class="stub"1
a2n-1a2n
)
≤class="stub"5
24
-class="stub"1
9•23
+class="stub"1
9
(class="stub"1
24
+…+class="stub"1
2n
)
=class="stub"5
24
-class="stub"1
9•23
+class="stub"1
9
•class="stub"1
23
(1-class="stub"1
2n-3
)< class="stub"5
24

综上,当n∈N*时,class="stub"1
6
Tn≤class="stub"5
24

更多内容推荐