数列an的前n项和记为sn,已知a1=1,a［n＋1］=n＋2/n×sn,求正数列｛sn/n｝是等比
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a(n+1)= [(n+2)/n].Sn
S(n+1)-Sn =[(n+2)/n].Sn
nS(n+1) = 2(n+1)Sn
S(n+1)/(n+1) = 2[ Sn/n]
=>{ Sn/n} 是等比数梁念列,q=2
Sn/洞渣凯n - S1/1 = 2(n-1)
Sn/纳唤n = 2n-1
Sn = (2n-1)n
for n>=2
an = Sn -S(n-1)
=2(2n-1) -1
=4n -3
S(n+1)-Sn =[(n+2)/n].Sn
nS(n+1) = 2(n+1)Sn
S(n+1)/(n+1) = 2[ Sn/n]
=>{ Sn/n} 是等比数梁念列,q=2
Sn/洞渣凯n - S1/1 = 2(n-1)
Sn/纳唤n = 2n-1
Sn = (2n-1)n
for n>=2
an = Sn -S(n-1)
=2(2n-1) -1
=4n -3
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