问题 解答题
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].
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 )∵

1
xn+xn+1
=
1
2 (
n
+
n+1
)
=
1
2
(
n+1
-
n
)

1
x1+x 2
+
1
x2+x3
+…+
1
xn+xn+1
=
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].

单项选择题
多项选择题