求一數列 ?
設有一數列<an>
滿足a1=6,a2=11
2an+1+3an
an+2=--------------------------
5
試求(1)an (2)lim an
n→∞
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請按下補充裡面的連結....上面的連結有問題...
3 個解答
- popi_popiLv 51 0 年前
題目:設一數列<a(n)>,a(1)=6,a(2)=11,a(n+2)=[2*a(n+1)+3*a(n)]/5
求(1) a(n)=?
(2)lim(n→∞) a(n)=?
sol:
(1)
(!)因a(n+2)=[2*a(n+1)+3*a(n)]/5
5*a(n+2)=2*a(n+1)+3*a(n)
5*a(n+2)-5*a(n+1)=-5*a(n+1)+2*a(n+1)+3*a(n)
化成5*[a(n+2)-a(n+1)]=-3*[a(n+1)-a(n)]
==>a(n+2)-a(n+1)=(-3/5)*[a(n+1)-a(n)]..........等比型式
(!!)上式首項=a(3)-a(2)=(-3/5)*(a(2)-a(1))=-3,公比=-3/5
所以第n-2項=a(n)-a(n-1)=(-3)*(-3/5)^(n-3)=5*(-3/5)^(n-2)
或另思考:a(n+2)-a(n+1)=(-3/5)*[a(n+1)-a(n)]=(-3/5)^2*[a(n)-a(n-1)]
=....=(-3/5)^n*[a(2)-a(1)]
==>a(n)-a(n-1)=5*(-3/5)^(n-2)
(!!!)
a(2)-a(1)=5*(-3/5)^0
a(3)-a(2)=5*(-3/5)^1
.......
a(n)-a(n-1)=5*(-3/5)^(n-2)
各式相加
a(n)-a(1)=5*[1+(-3/5)^1+(-3/5)^2+......+(-3/5)^(n-2)]
a(n)=6+5*[1-(-3/5)^(n-1)]/[1-(-3/5)]=(73/8)-(25/8)*(-3/5)^(n-1)
(2)lim(n→∞) a(n)=lim(n→∞) (73/8)-(25/8)*(-3/5)^(n-1)=73/8
2009-12-16 16:58:54 補充：
考慮一下
還是答了~
參考之~
2009-12-16 17:05:28 補充：
補充:
設p*a(n+2)+q*a(n+1)+r*a(n)=0,且p+q+r=0
==>a(n)=a(1)+[a(2)-a(1)]*{[1-(r/p)^(n-1)]/[1-(r/p)]}
- 浮浪貢Lv 71 0 年前
1.
5a(n+2)= 2a (n+1)+ 3a (n)……
整理成…… (a(n+2) +B a(n+1) )=A((a(n+1) +B a(n) )
比較係數得B=3/5,A=1
2.
辗轉代入得
(a(n) +B a(n-1) )=(a(n-1) +B a(n-2) )= (a(n-2) +B a(n-3) )
=………………..=(a(2) +B a(1) )
= (11+B*6)
= (73/5)
a(n) =- B a(n-1)+73/5
3.
a(n) =(-3/5) a(n-1) + (73/5)…….辗轉代入得
=(-3/5)[ (-3/5) a(n-2) + (73/5)]+ (73/5)
=(-3/5)^2 *a(n-2) + (73/5) (-3/5) + (73/5)
=……..
=(-3/5)^(n-2) *a(2) + (73/5)[ (-3/5)^(n-3)+ (-1/5)^(n-4)+…+1]
=(-3/5)^(n-2) *11 +(73/5)[1-(-3/5)^(n-2)]/[1-(-3/5)]
=(-3/5)^(n-2) *11+73/8-(73/8) (-3/5)^(n-2)
4. lim a(n) =73/8
