已知数列{an}满足nan+1=(n+1)an+2n(n+1),n属于正整数,且a1=1,设bn=an+n.（1）求｛an｝；（2）设n大于等于2,S=1/b2-2+1/b3-2+...+1/bn-2,求S的值关键是第二问啊,第一问我会了.

Question

已知数列{an}满足nan+1=(n+1)an+2n(n+1),n属于正整数 ,且a1=1,设bn=an+n.
（1）求｛an｝；（2）设n大于等于2,S=1/b2-2 + 1/b3-2 + ...+1/bn-2 ,求S的值
关键是第二问啊,第一问我会了.

Answer 1

1.
等式两边同除以n(n+1)
a(n+1)/(n+1)=an/n +2
a(n+1)/(n+1) -an/n=2,为定值.
a1/1=1/1=1
数列{an/n}是以1为首项,2为公差的等差数列.
an/n=1+2(n-1)=2n-1
an=n(2n-1)=2n²-n
2.
bn=an+n=2n²-n+n=2n²
1/(bn -2)=1/(2n²-2)=(1/2)[1/(n²-1)]=(1/2)[1/(n+1)(n-1)]=(1/2)[1/(n-1)-1/(n+1)]
Sn=1/(b2-2)+1/(b3-2)+...+1/(bn-2)
=(1/2)[1-1/3+1/2-1/4+...+1/(n-1)-1/(n+1)]
=(1/2)[ [1+1/2+1/3+...+1/(n-1)]-[1/3+1/4+...+1/(n+1)] ]
=(1/2)[1+1/2 -1/n -1/(n+1)]
=3/4 -1/(2n) -1/[2(n+1)]