n=1時(shí),3a1=3a1,n=2時(shí),3+3a2=4a2,a2=3
3（a1+a2+a3+······+an）=（n+2）an①
n>=2時(shí)有：
3（a1+a2+a3+······+a(n-1)）=（n+1）a(n-1)②
①-②得3an=(n+2)an-(n+1)a(n-1)
化簡(jiǎn)得(n-1)an=(n+1)a(n-1)即an/a(n-1)=(n+1)/(n-1)
.
a4/a3=5/3
a3/a2=4/2
a2/a1=3/1
以上各式相乘,消去相同的項(xiàng),an/a1=n*(n+1)/(1*2) 且 a1=1
所以an=n*(n+1)/2
n=1時(shí),a1=1*2/2=1也成立,所以an=n*(n+1)/2