设数列an的前n项和为sn 若对于任意的正整数都有Sn=2an-3n

S1=a1
令n=1有：
S1=2a1-3
a1=3
S[n+1]=S[n]+a[n+1]
S[n+1]=2a[n+1]-3(n+1)
S[n]=2a[n]-3n
两式相减有：
S[n+1]-S[n]=2a[n+1]-2a[n]-3
即
a[n+1]=2a[n+1]-2a[n]-3
所以有：
a[n+1]=2a[n]+3