Question

这道数列题怎样做？急求

Answer 1

（1）解:
∵﹛an﹜为等差数列
∴设an=bn+m,则an+1=b(n+1)+m=bn+b+m
所以有bn+b+m=(bn+m)²-n(bn+m)+1
∴bn+b+m=b²n²+2bmn+m²-bn²-mn+1
∴(b²-b)n²+(2bm-m-b)n+m²-m-b+1=0
∵此方程对任意n∈N恒成立
∴b²-b=0,2bm-m-b=0,m²-m-b+1=0
∴b=1,m=1
∴an=n+1
(2)证明:
1/a(n+1)+1/a(n+2)+...+1/a(2n+1)
=1/(n+2)+1/(n+3)+....+1/(2n+2)
设f(n)=1/(n+2)+1/(n+3)+....+1/(2n+2)
f(n+1)-f(n)=[1/(n+3)+1/(n+4)+....+1/(2n+3)]-[1/(n+2)+1/(n+3)+....+1/(2n+2)]
=1/(2n+3)-1/(n+2)
=-(n+1)/(2n+3)(n+2)＜0
∴f(n)是一个单调递减的函数
∴f(n)≤f(1)=1/3＜ln2
∴1/a(n+1)+1/a(n+2)+...+1/a(2n+1)＜ln2

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