3,4,5.5,12,13.7,24,25.9,40,41....a,b
问题描述：
3,4,5.5,12,13.7,24,25.9,40,41....a,b
3,4,5.5,12,13.7,24,25.9,40,41....a,b,c.
当a等于19时求b、c的值.
当a等于2n加一时,求b、c的值.
用（2）的结论判断15,111,112是否为一组勾股数并说明理由.
答
（1）发现4 = 3*2 – 2,5 = 4 + 1 ；
12 = 5*3 – 3,13 = 12 + 1 ；
24 = 7*4 – 4,25 = 24 + 1 ；
40 = 9*5 – 5,41 = 40 + 1 ；
勾股数a2 + b2 = c2 ；
所以当a = 19时,b = 19*10 – 10 = 180,c = b + 1 = 180 + 1 = 181 ；
综上所述,b = 180,c = 181 ；
（2）当a = 2n + 1时,b = (2n + 1)*(n + 1) – (n + 1) = 2n2 + 3n + 1 – n – 1 = 2n2 + 2n,c = b + 1 = 2n2 + 2n + 1 ；
验算：a2 + b2 = (2n + 1)2 + (2n2 + 2n)2 = 4n2 + 4n + 1 + 4n4 + 8n3 + 4n2 = 4n4 + 8n3 + 8n2 + 4n + 1 = (2n2 + 2n + 1)2 = c2,符合题意.
综上所述,b = 2n2 + 2n,c = 2n2 + 2n + 1 .