• 2022-07-01
    求数列[img=164x46]1803072d931eae3.png[/img]的通项公式
    A: RSolve[{a[n+1]==(2a[n]+3)/(a[n]+4),a[0]==0},a[n],n]
    B: RSolve[a[n+1]==(2a[n]+3)/(a[n]+4),a[0]==0,a[n],n]
    C: RSolve[{a[n+1]==(2a[n]+3)/(a[n]+4),a[0]==0},a[n]]
    D: RSolve[{a_(n+1)==(2a_n+3)/(a_n+4),a_0==0},a_n,n]