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)
證明上述猜想 .
數(shù)學(xué)歸納法:
1# n =1時(shí),a1=1*2; b1=(1+1)^2 = 4 成立
2# 假設(shè) n = k; 有 ak = k(k+1) ;bk = (k+1)^2
n=k+1 時(shí) 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
所以證明了 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)