.已知数列的前n项之和为Sn=n2+3n,求证{an}为等差数列,若Sn=n2+3n+1呢?

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.已知数列的前n项之和为Sn=n2+3n,求证{an}为等差数列,若Sn=n2+3n+1呢?
.已知数列的前n项之和为Sn=n2+3n,求证{an}为等差数列,若Sn=n2+3n+1呢?

.已知数列的前n项之和为Sn=n2+3n,求证{an}为等差数列,若Sn=n2+3n+1呢?
由Sn=n^2+3n得
S(n-1)=(n-1)^2+3(n-1),两式相减,考虑到Sn-S(n-1)=an得
an=2n-1+3=2n+4,于是得
a(n-1)=2(n-1)+4,两式相减得
an-a(n-1)=2,故{an}为等差数列.
如果Sn=n2+3n+1,则
S(n-1)=(n-1)^2+3(n-1)+1,两式相减得
an-a(n-1)=2,故{an}仍为等差数列.