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
(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=2n
∵x1=2也满足上式,∴xn=2n
(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
)=31
∴n=48
(3)xnxn+1=2
2n
=4n+1 n
<4n+1
=4n+2n+(n+1) 2
∴x1x2+x2x3+…xnxn+1<(4×1+2)+(4×2+2)+…(4n+2)=
n=2[(n+1)2-1].6+(4n+2) 2