# 求一數列 ?

2an+1+3an

an+2=--------------------------

5

n&rarr;&infin;

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2 個已更新項目:

### 3 個解答

• 筱紫
Lv 7
1 0 年前
最佳解答

有錯請指正, 第一題與樓上大大略有不同!

若看不到請點連結

http://blog.xuite.net/ginwha/school/29343222

• 1 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)]}

• 1 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