Question

在数列{an},{bn}中，a1=2,b1=4,且an，bn，a(n+1)成等差数列，bn,a(n+1),b(n+1)成等比数列（n€n*)

1）求a2，a3，a4及b1，b2，b3，由此猜测{an},{bn}的通项公式；
2）证明：1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+~~~+1/(an+bn)<5/12。

Answer 1

<1>可知 a2 = 6,b2 = 9,a3=12,b3=16;a4=20
猜想 an = a(n-1) +2*n ; bn = (n+1)^2
可知 an = 2(1+2+..+n) = n(n+1)

<2> 证明上述猜想 。
数学归纳法:
1# n =1时, a1=1*2; b1=(1+1)^2 = 4 成立
2# 假设 n = k; 有 ak = k(k+1) ;bk = (k+1)^2
n=k+1 时 a(k+1) = 2bk-ak = 2k^2+4k+2-k^2-k = k^2+3k+2=(k+1)(k+2)
b(k+1) = [a(k+1)]^2/bk = (k+1)^2(k+2)^2/(k+1)^2 = (k+2)^2 也成立

所以证明了 an = (n+1)n ;bn=(n+1)^2
an+bn = (n+1)(2n+1) = (2n+2)(2n+1)/2
1/(an+bn) = 1/(n+1)(2n+1)<1/2n(n+1) = 1/2*(1/n-1/(n+1))

于是 1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+~~~+1/(an+bn)
= 2/(2*1+1)(2*1+2) + ....+ 2/(2n+1)(2n+2) [裂项相消]
<2(1/3-1/4)+1/2(1/2-1/3+1/3-1/4+.....+1/n-1/(n+1))
<1/6+1/4 = 10/24 = 5/12