已知数列an的前n项和为sn,a1=2,nan+1=sn+n(n+1),设bn=1⼀ana（n+1）a（n+2），证明b1+b2+b3……+bn<1⼀32

na(n+1) = s(n) + n(n+1) = n[s(n+1) - s(n)],
ns(n+1)=(n+1)s(n) + n(n+1),
s(n+1)/(n+1) = s(n)/n + 1,
{s(n)/n}是首项为s(1)/1 = a(1) = 2,公差为1的等差数列。
s(n)/n = 2 + (n-1) = n+1,
s(n) = n(n+1).

na(n+1) = s(n) + n(n+1) = 2n(n+1),
a(n+1) = 2(n+1).
又，a(1)=2,
因此，总有，
a(n) = 2n.

b(n) = 1/[a(n)a(n+1)a(n+2)] = 1/[8n(n+1)(n+2)]
= 1/[16n(n+1)] - 1/[16(n+1)(n+2)]

b(1)+b(2)+...+b(n-1)+b(n)
= 1/[16*1*2] - 1/[16*2*3] + 1/[16*2*3] - 1/[16*3*4] + ... + 1/[16(n-1)n] - 1/[16n(n+1)] + 1/[16n(n+1)] - 1/[16*(n+1)(n+2)]
= 1/[16*1*2] - 1/[16(n+1)(n+2)]
< 1/[16*1*2]
= 1/32