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)
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/04 01:50:13
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(
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=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=2n²+2n我求出xn=2√n,如何证明x1x2+x2x3+x3x4……+xnx(n+1)
(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].