Question

设数列{an}满足a1=2,a(n+1)=(n/n+2)an(n∈N+)

(1)求这个数列的通项公式（2）求数列{an}的前n项和Sn

Answer 1

a(n+1)=(n/n+2)an
a(n)=(n-1)/(n+1)an-1
...
a2=1/(1+2)a1
上述等式左右分别累乘得
an+1=4/(n+1)(n+2)=4/(n+1)-4/(n+2)
sn=4-1/(n+1)

Answer 2

a(n)=(n-1)/(n+1)a（n-1）
a(n-1)=(n-2)/n*a(n-2)
...
a2=1/(1+2)a1
上述等式左右分别累乘得
an=4/(n(n+1))=4/n-4/(n+1)
sn=4-4/(n+1)=4n/(n+1)

Answer 3

a1=1,a2=1/3,a3=1*2/3*4 ,……,an=2(n-1)!/(n+1)!=2/n(n+1)
2、an=2(1/n-1/(n+1)),n>1
sn=1+2(1/2-1/3+1/3-1/4+……+1/n-1/(n+1))=2-2/(1+n)=2n/(n+1)