高一數學遞回關係

令(An + k) = 3(An-1 + k) ........(1)
An + k = 3An-1 + 3k
An = 3An-1 + 2k
上式和已知的 An = 3An-1 + 1 對照
得 2k = 1 → k = 1/2
代回(1)得
(An + 1/2) = 3(An-1 + 1/2) .......(2)

利用A1 = 1 和 (2)式
A1 + 1/2 = 1 + 1/2 = 3/2
A2 + 1/2 = 3(A1 + 1/2)
A3 + 1/2 = 3(A2 + 1/2)
A4 + 1/2 = 3(A3 + 1/2)
A5 + 1/2 = 3(A4 + 1/2)
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以上各式等號左邊的所有項相乘 = 等號右邊的所有項相乘
則相同顏色的都可以相消，剩下
A5 + 1/2 = (3/2)*3*3*3*3 = 243/2

所以A5 = 243/2 - 1/2 = 242/2 = 121

^____^

a1=1，an=3*a(n-1)+1，求a5=?Ans:基本理論: d,r=常數An=A(n-1)+d=等差級數.....d=公差An=r*A(n-1)=等比級數.....r=公比An=r*A(n-1)+d=特殊等比級數舉A4為例:A4=r*A3+d=r*(r*A2+d)+d=r^2*A2+d*(r+1)=r^2*(r*A1+d)+d*(r+1)=r^3*A1+d*(r^2+r+1)=r^3*A1+d*(r^3-1)/(r-1)=> An=r^m*A1+d*(r^m-1)/(r-1).....m=n-1r=公比=3, d=公差=1, A1=首項=1A5=r^4*A1+d*(r^4-1)/(r-1)=81*1+1*(81-1)/(3-1)=81+80/2=81+40=121.....ansA6=243+(243-1)/2=243+121=364.....

2014-01-26 07:52:39 補充：
基本理論補充:

An=A(n-1)+d=等差數列.....d=公差

An=r*A(n-1)=等比數列.....r=公比

An=r*A(n-1)+d=x*等比數列+y*等差數列.....x,y=待定=?

等差數列=F(n)

F(n)=常數=1階等差數列

F(n)=1次式=2階等差數列

F(n)=2次式=3階等差數列

.......

以an=3*a(n-1)+1=x*3^m+y為例: m=n-1, r=3, d=1

a1=3*a0+1.....a0=0

=1

=x*3^m+y.....m=n-1=0

=x+y => x+y=1

2014-01-26 07:53:07 補充：
a2=3*a1+1

=4

=x*3+y => 3x+y=4

相減: 2x=3 => x=3/2, y=-1/2

=> an=3*3^m/2-1/2=(3^n-1)/2.....ans

a6=(3^6-1)/2

=(27*27-1)/2

=(729-1)/2

=364.....ans

2014-01-26 08:10:47 補充：
雙變數an,a(n-1): F(n)=常數

3變數an,a(n-1),a(n-2): F(n)=b*n+c

......

{an}
= 3{an-1}+1
= 3(3{an-2}+1)+1
= 9{an-2}+3+1
= 9(3{an-3}+1)+3+1
= 27{an-3}+9+3+1
= ...
= 3^(n-1)*{a1}+ [3^(n-1)-1]/2

{a5}
= 3^4*{a1} + [3^4-1]/2
= 81*1+40
= 121

a(n)=3 a(n-1)+1，求a5

令
a(n)-x=3 [ a(n-1) - x ]
a(n)=3 a(n-1) - 2x ==> -2x = 1 , x=-1/2
a(n) + 1/2 =3 [ a(n-1) +1/2 ]

a(5) +1/2 =3^4 [ a(1) + 1/2 ]=81( 1+1/2)=243 / 2

a(5) =243 /2 - 1/2 =121

2014-01-24 23:28:29 補充：
a1=1，an=3an-1+1，求a5
一個一個推也行
a2 = 3*1+1=4
a3=3*4+1=13
a4=3*13+1=40
a5=40*3+1=121