數列：1.2.3.5.8.13.21.34......有無通式
問題為.後面的數字是前面兩項數字的和
求特定數列的大小...例如第86項的數字為何
2005-04-16 23:06:35 · 5 個解答 · 發問者 Sifes 2 in 教育與參考 ➔ 其他：教育
這跟斐波那契的數字(Fibonacci Number)差不多. 只不過斐波那契的數字(Fibonacci Number)多一個1在前面
1, 2, 3, 5, 8, 13, 21, 34, …
1, 1, 2, 3, 5, 8, 13, 21, 34, …
也就是說f(x) = F(x+1). F是斐波那契(Fibonacci)的函數. f是你數字串的函數
斐波那契的數字(Fibonacci Number)的公式是
(POWER((1+POWER(5, 0.5))/2, A1)-POWER((1-POWER(5, 0.5))/2, A1))/POWER(5, 0.5)
F(X) = ((((1+5^0.5)/2)^X)-(((1-5^0.5)/2)^X))/(5^0.5)
所以你要的公式是
f(X) = ((((1+5^0.5)/2)^(X+1))-(((1-5^0.5)/2)^(X+1)))/(5^0.5)
f(1) = 1
f(2) = 2
f(3) = 3
f(4) = 5
...
f(86) = 679891637638612000
P. S. 第86個後三個數字並不是0. 我的計算機(excel)不夠精準所以後三個數字顯示0出來
懂不懂?不懂請再問
2005-04-18 01:31:12 補充：
不用客氣... :)
2005-04-17 01:09:55 · answer #1 · answered by 龍 7 · 0⤊ 0⤋
Fibonacci 的標準定義，是 F(0) =0, F(1) = 1, F(n) = F(n-1) + F(n-2)
2007-04-19 20:53:48 · answer #2 · answered by ? 7 · 0⤊ 0⤋
這是我弟在補習班看到的類似題目...原題目是第24項為多少...我弟來問我這怎麼算..我說加一加吧...弟反問我有無通式...我就自己導了一些式子...是有發現規則...不過無法變為通式...故上網求助...對此感謝回答我題目的各為大大...感謝萬分ORZ...將奉上分數以為感謝
2005-04-17 21:04:24 · answer #3 · answered by Sifes 2 · 0⤊ 0⤋
fibonacci數列第n項的為
F_n = ( (1+√5)^n - (1-√5)^n ) / ( 2^n * √5 )
你要求的第86項是
F_85 = ( (1+√5)^86 - (1-√5)^86 ) / ( 2^86 * √5 )
2005-04-17 01:24:21 · answer #4 · answered by ? 6 · 0⤊ 0⤋
會死人ㄉ>"<
附上我推出來A1-A100ㄉ數字
a11 a26196418 a5132951280099 a765527939700884760
a22 a27317811 a5253316291173 a778944394323791460
a33 a28514229 a5386267571272 a7814472334024676200
a45 a29832040 a54139583862445 a7923416728348467700
a58 a301346269 a55225851433717 a8037889062373143900
a613 a312178309 a56365435296162 a8161305790721611600
a721 a323524578 a57591286729879 a8299194853094755500
a834 a335702887 a58956722026041 a83160500643816367000
a955 a349227465 a591548008755920 a84259695496911123000
a1089 a3514930352 a602504730781961 a85420196140727490000
a11144 a3624157817 a614052739537881 a86679891637638612000
a12233 a3739088169 a626557470319842 a871100087778366100000
a13377 a3863245986 a6310610209857723 a881779979416004710000
a14610 a39102334155 a6417167680177565 a892880067194370820000
a15987 a40165580141 a6527777890035288 a904660046610375530000
a161597 a41267914296 a6644945570212853 a917540113804746350000
a172584 a42433494437 a6772723460248141 a9212200160415121900000
a184181 a43701408733 a68117669030460994 a9319740274219868200000
a196765 a441134903170 a69190392490709135 a9431940434634990100000
a2010946 a451836311903 a70308061521170129 a9551680708854858300000
a2117711 a462971215073 a71498454011879264 a9683621143489848400000
a2228657 a474807526976 a72806515533049393 a97135301852344707000000
a2346368 a487778742049 a731304969544928660 a98218922995834555000000
a2475025 a4912586269025 a742111485077978050 a99354224848179262000000
a25121393 a5020365011074 a753416454622906710 a100573147844013817000000
a0=1
a1=2
a2=a0+a1
a3=a1+a2
a4=a2+a3=2a2+a1
a5=a3+a4=3a2+2a1
a6=a4+a5=5a2+3a1
a7=a5+a6=8a2+5a1
a8=a6+a7=13a2+8a1
a9=a7+a8=21a2+13a1
a10=a8+a9=34a2+21a1
a11=a9+a10=55a2+34a1
a12=a10+a11=89a2+55a1
最元始的推法........不過會死人
我的算法是...假定a0=1 a1=2 a3=3 a4=5....
我是用等差級數的觀念去看
上面的數據是我用EXCEL推的
不過....基本上
我贊成2樓大大說的
這個數字......你用迴圈算....有學過QB 或VB的就會知道的
比較方便
2005-04-16 23:37:20 · answer #5 · answered by 心鏡 2 · 0⤊ 0⤋