设数列【an】满足a1=1,3（a1+a2+a3+······+an）=（n+2）an,求通项an

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设数列【an】满足a1=1,3（a1+a2+a3+······+an）=（n+2）an,求通项an
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设数列【an】满足a1=1,3（a1+a2+a3+······+an）=（n+2）an,求通项an
设数列【an】满足a1=1,3（a1+a2+a3+······+an）=（n+2）an,求通项an

设数列【an】满足a1=1,3（a1+a2+a3+······+an）=（n+2）an,求通项an
n=1时,3a1=3a1,n=2时,3+3a2=4a2,a2=3
3（a1+a2+a3+······+an）=（n+2）an①
n>=2时有：
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)
化简得(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,消去相同的项,an/a1=n*(n+1)/(1*2) 且 a1=1
所以an=n*(n+1)/2
n=1时,a1=1*2/2=1也成立,所以an=n*(n+1)/2

1:3(a1+a2+a3+.....+an)=(n+2)an
2:3(a1+a2+a3+.....+a(n+1))=(n+1+2)a(n+1)
2-1=>3a(n+1)=(n+3)a(n+1)-(n+2)an=>(n+2)an=na(n+1)=>a(n+1)=(n+2)/n*an=(n+2)/n*(n+1)/(n-1)*......3/1*a1=>(n+2)*(n+1)/2
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用累商法去解题

a1+a2+a3+.....+an=sn
3sn=(n+2)an
3sn-1=(n+1)an-1(n≥2)
3sn-3sn-1=(n+2)an-(n+1)an-1=3an
a2/a1=3/1,a3/a2=4/2 ....连乘有an/a1=n(n+1)/2
a1=1
所以an=n(n+1)/2, n=1时成立

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