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*)
问题描述：
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)
答
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)
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)