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
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
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
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