Question

已知数列{an}的前n项和为Sn,若a1=2,an+1=Sn+n(n+1),(1)证明数列{Sn/n}为等差数列，并求Sn

Answer 1

s(1)=a(1)=2,
s(n)+n(n+1)=a(n+1)=s(n+1)-s(n),
s(n+1)=2s(n)+n(n+1),
s(n+1)+n(n+1)=2s(n)+2n(n+1)=2s(n)+2n(n-1)+4n,
s(n+1)+n(n+1)+4(n+1)=2[s(n)+(n-1)n+4n]+4,
s(n+1)+n(n+1)+4(n+1)+4=2[s(n)+(n-1)n+4n+4]
{s(n)+(n-1)n+4(n+1)}是首项为s(1)+8=10,公比为2的等比数列.
s(n)+(n-1)n+4(n+1)=10*2^(n-1),
s(n)=10*2^(n-1) - n(n-1) - 4(n+1),

s(n)/n不是等差数列哈...