设数列{xn}各项为正,且满足x1²+x2²+x3²+……+xn²=2n²+2n

题目

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)

答案

（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）x_nx_n+1=2
n
2
n+1
=4
nn+1
＜4
n+(n+1)2
=4n+2
∴x₁x₂+x₂x₃+…x_nx_n+1＜（4×1+2）+（4×2+2）+…（4n+2）=
6+(4n+2)2
n=2[（n+1）²-1]．