数列<an>的前n项和为Sn,a1=1,且nSn+1=(n+2)Sn(n∈N☆)
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nS(n+1)=(n+2)Sn
S(n+1)/Sn = (n+2)/n
Sn/S(n-1) = (n+1)/(n-1)
Sn/S1 = (n+1)n/2
Sn = (n+1)n/2
an = Sn -S(n-1)
= n
