设数列{xn}各项为正,且满足x12+x22+…xn2=2n2+2n.(1)求xn;(2)已知1x1+x2+1x2+x3+…+1xn+xn+1=3,求n;(3)证明:x1x2+x2x3+…xnxn+1

题目简介

设数列{xn}各项为正,且满足x12+x22+…xn2=2n2+2n.(1)求xn;(2)已知1x1+x2+1x2+x3+…+1xn+xn+1=3,求n;(3)证明:x1x2+x2x3+…xnxn+1

题目详情

设数列{xn}各项为正,且满足x12+x22+…xn2=2n2+2n.
(1)求xn
(2)已知
1
x1+x 2
+
1
x2+x3
+…+
1
xn+xn+1
=3
,求n;
(3)证明:x1x2+x2x3+…xnxn+1<2[(n+1)2-1].
题型:解答题难度:中档来源:不详

答案

(1)∵数列{xn}各项为正,且满足x12+x22+…xn2=2n2+2n.
∴x1=2
当n,xn2=2n2+2n-[2(n-1)2+2(n-1)]=4n,∴xn=2
n

∵x1=2也满足上式,∴xn=2
n

(2 )∵class="stub"1
xn+xn+1
=class="stub"1
2 (
n
+
n+1
)
=class="stub"1
2
(
n+1
-
n
)

class="stub"1
x1+x 2
+class="stub"1
x2+x3
+…+class="stub"1
xn+xn+1
=class="stub"1
2
(
n+1
-
1
)
=3
∴n=48
(3)xnxn+1=2
n
2
n+1
=4
n
n+1
4
n+(n+1)
2
=4n+2
∴x1x2+x2x3+…xnxn+1<(4×1+2)+(4×2+2)+…(4n+2)=
6+(4n+2)
2
n
=2[(n+1)2-1].

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