a(n+1)+an=r*2^(n-1)1
a(n+1)-p*an=p*t 2
a1=1
想要求出p,r,t,只需要分別求出兩個(gè)數(shù)列的通項(xiàng),使其對(duì)應(yīng)參數(shù)相等即可
對(duì)于數(shù)列1：
假定a(n+1)+an=r(m*2^(n+1)+m*2^n)
則3*m=1/2,m=1/6
a(n+1)-(r/6)*2^(n+1)=(-1)(an-(r/6)*2^n)
a(n)=(-1)^(n-1)(a1-r/3)+(r/6)*2^n
對(duì)于數(shù)列2：
假定a(n+1)-p*an=p*(k*t-pk*t)
則k=1/(1-p),
a(n+1)-p*t/(1-p)=p*(an-p*t/(1-p))
an=p^(n-1)(a1-p*t/(1-p))+p*t/(1-p)
比較兩式,由于r≠0,則數(shù)列2常數(shù)項(xiàng)必須為0
根據(jù)通項(xiàng),p≠0,得出t=0,可進(jìn)一步推出p=2,r=3
代入得到,an=2^(n-1)