题目内容：

设公差不为0的等差数列an的首项为1,且a6=a2^2,求数列an的通项公式,若bn满足bn=nan+5n,求{1/bn}的前n项和Tn

优质解答

an=a1+(n-1)da1=1a6= (a2)^2(1+5d) = (1+d)^2d^2-3d=0d=3an = 1+3(n-1) = 3n-2bn=nan+5n= n(3n-2) + 5n= 3n(n+1)1/bn = 1/[3n(n+1)]= (1/3)[1/n -1/(n+1) ]Tn = 1/b1+1/b2+...+1/bn= (1/3)[1-1/(n+1) ]=n/[3(n+1)]