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设数列{xn}各项为正,且满足x1²+x2²+x3²+……+xn²=2n

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设数列{xn}各项为正,且满足x1²+x2²+x3²+……+xn²=2n²+2n
我求出xn=2√n,如何证明x1x2+x2x3+x3x4……+xnx(n+1)
设数列{xn}各项为正,且满足x1²+x2²+x3²+……+xn²=2n
(2 )∵
1xn+xn+1
=
12 (n+n+1)
=
12
(
n+1

n
)

1x1+x 2
+
1x2+x3
+…+
1xn+xn+1
=
12
(
n+1

1
)=3
∴n=48
(3)xnxn+1=2
n
2
n+1
=4
nn+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].