數列

光武

1,5,13,27,48,78,118,170 ...

a)求以上數列的通項 T(n)
b)求第75項.

唔該x100

98102257

條數係中四, 中五的教課書搵出來 ?
點解 d 數字好似無咩關連

光武

原帖由 98102257 於 2009-3-31 08:53 PM 發表
條數係中四, 中五的教課書搵出來 ?
點解 d 數字好似無咩關連

條數系個牙SIR出比我..我諗到D數字既關系 旦系搵唔到個通項
應該唔算系 等差/等比 數列

[ 本帖最後由 光武 於 2009-4-1 10:10 PM 編輯 ]

98102257

原帖由 光武 於 2009-4-1 10:08 PM 發表

條數系個牙SIR出比我..我諗到D數字既關系 旦系搵唔到個通項
應該唔算系 等差/等比 數列

可唔可以講下 d 數字有咩關係
等我同你諗下個通項

98102257

唔好見我打左咁多字以為條數好難...其實難就唔係..不過好煩囉
同埋條數好講求準確性...只要有一步計錯...就好大獲
條數我都搞左成個半鐘...因為過程中我成日計錯
不過最後我搞到個通項...我都好開心
你慢慢研究下啦
可能仲有更好的方法...所以你參考下好了
---------------------------------------------------------------------------

T(1)=1
T(2)=5
T(3)=13
T(4)=27
T(5)=48
T(6)=78
T(7)=118
T(8)=170
...

T(2)-T(1)=4(1)
T(3)-T(2)=4(2)
T(4)-T(3)=7(2)
T(5)-T(4)=7(3)
T(6)-T(5)=10(3)
T(7)-T(6)=10(4)
T(8)-T(7)=13(4)
...

LET S(n)=T(n)-T(n-1) ... (A)
S(2)=4(1)
S(3)=4(2)
S(4)=7(2)
S(5)=7(3)
S(6)=10(3)
S(7)=10(4)
S(8)=13(4)
...

LET P(n)=S(n)-S(n-1) ... (B)
P(3)=4(1)
P(4)=3(2)
P(5)=7(1)
P(6)=3(3)
P(7)=10(1)
P(8)=3(4)
...

LET Q(n)=P(2n+3)-P(2n+1) ... (C)
Q(1)=P(5)-P(3)=3
Q(2)=P(7)-P(5)=3
...
Q(n)=3

LET F(n)=P(2n+4)-P(2n+2) ... (D)
F(1)=P(6)-P(4)=3
F(2)=P(8)-P(6)=3
...
F(n)=3

Consider Q(1)+Q(2)+...+Q(n)
=[P(5)-P(3)]+[P(7)-P(5)]+...+[P(2n+3)-P(2n+1)]
=P(2n+3)-P(3)
=P(2n+3)-4
On the other hand, Q(1)+Q(2)+...+Q(n)=3n
So 3n=P(2n+3)-4
P(2n+3)=3n+4 ... (E)

Consider F(1)+F(2)+...+F(n)
=[P(6)-P(4)]+[P(8)-P(6)]+...+[P(2n+4)-P(2n+2)]
=P(2n+4)-P(4)
=P(2n+4)-6
On the other hand, F(1)+F(2)+...+F(n)=3n
So 3n=P(2n+4)-6
P(2n+4)=3n+6 ... (F)

Consider P(3)+P(4)+...+P(2n+4)
=[S(3)-S(2)]+[S(4)-S(3)]+...+[S(2n+4)-S(2n+3)]
=S(2n+4)-S(2)
=S(2n+4)-4
On the other hand, P(3)+P(4)+...+P(2n+4)
=P(3)+P(4)+[P(5)+P(7)+...+P(2n+3)]+[P(6)+P(8)+...+P(2n+4)]
=4+6+[3(1+2+...+n)+4n]+[3(1+2+...+n)+6n]  By (E) & (F)
=3n^2+13n+10
So S(2n+4)-4=3n^2+13n+10
S(2n+4)=3n^2+13n+14 ... (G)

Since P(2n+4)=S(2n+4)-S(2n+3)
So S(2n+3)=S(2n+4)-P(2n+4)=(3n^2+13n+14)-(3n+6)=3n^2+10n+8 ... (H)

Consider S(2)+S(3)+...+S(2n+4)
=[T(2)-T(1)]+[T(3)-T(2)]+...+[T(2n+4)-T(2n+3)]
=T(2n+4)-T(1)
=T(2n+4)-1
On the other hand, S(2)+S(3)+...+S(2n+4)
=S(2)+S(3)+S(4)+[S(6)+S(8)+...+S(2n+4)]+[S(5)+S(7)+S(9)+...+S(2n+3)]
=4+8+14+[3(1^2+2^2+...+n^2)+13(1+2+...+n)+14n]+[3(1^2+2^2+...+n^2)+10(1+2+...+n)+8n)  By (G) & (H)
=2n^3+(29/2)n^2+(69/2)n+26
So T(2n+4)-1=2n^3+(29/2)n^2+(69/2)n+26
T(2n+4)=2n^3+(29/2)n^2+(69/2)n+27 ... (I)

Since S(2n+4)=T(2n+4)-T(2n+3)
So T(2n+3)=T(2n+4)-S(2n+4)=(2n^3+(29/2)n^2+(69/2)n+27)-(3n^2+13n+14) By (G) & (I)
T(2n+3)=2n^3+(23/2)n^2+(43/2)n+13 ...(J)

T(75)
Put n=36 in (J), then we can find the value.
通項
T(2n+3)=2n^3+(23/2)n^2+(43/2)n+13
T(2n+4)=2n^3+(29/2)n^2+(69/2)n+27

Checking
n=-1
T(1)=-2+(23/2)-(43/2)+13=1
T(2)=-2+(29/2)-(69/2)+27=5
n=0
T(3)=13
T(4)=27
...
應該無問題

[ 本帖最後由 98102257 於 2009-4-2 02:28 AM 編輯 ]

光武

強 & thxx

Sikkwong

原帖由 98102257 於 2009-4-2 02:25 AM 發表
唔好見我打左咁多字以為條數好難...其實難就唔係..不過好煩囉
同埋條數好講求準確性...只要有一步計錯...就好大獲
條數我都搞左成個半鐘...因為過程中我成日計錯
不過最後我搞到個通項...我都好開心
你慢慢研究下 ...

邊有咁簡單= =你個ans錯

通項=[n(n+2)(2n+1)]/8<-----for n even
        [n(n+2)(2n+1)-1]/8 <---for n odd
步驟根本唔係中4,5野

98102257

原帖由 Sikkwong 於 2009-4-22 06:34 PM 發表

邊有咁簡單= =你個ans錯

通項=[n(n+2)(2n+1)]/8

我錯都唔出奇
想問下你點計