已知an=5^[2^(n-1)] -3,设bn={1/（an-6）}-{1/（an²+6an）},数列bn的前n项和为Tn,
问题描述：
已知an=5^[2^(n-1)] -3,设bn={1/（an-6）}-{1/（an²+6an）},数列bn的前n项和为Tn,
求证-5/16≤Tn＜-1/4
答
这题看上去很吓人,但做起来不麻烦,别被吓住就好~构造数列cn ; 使 cn = an-6cn = an- 6 = 5^[2^(n-1)] -9an^2+6n = (an+3)^2- 9 = [5^[2^(n-1)]]^2 -9 = 5^[2^(n)] -9 = c(n+1) !也就是说 bn = 1/cn-1/c(n+1)于是 Tn ...