Math and Arithmetic

What are the first 100 numbers of the Fibonacci sequence?

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2012-02-22 07:10:49
2012-02-22 07:10:49

The first 101 numbers of the Fibonacci sequence, because zero is the critical base:

0

1

1

2

3

5

8

13

21

34

55

89

144

233

377

610

987

1597

2584

4181

6765

10946

17711

28657

46368

75025

121393

196418

317811

514229

832040

1346269

2178309

3524578

5702887

9227465

14930352

24157817

39088169

63245986

102334155

165580141

267914296

433494437

701408733

1134903170

1836311903

2971215073

4807526976

7778742049

12586269025

20365011074

32951280099

53316291173

86267571272

139583862445

225851433717

365435296162

591286729879

956722026041

1548008755920

2504730781961

4052739537881

6557470319842

10610209857723

17167680177565

27777890035288

44945570212853

72723460248141

117669030460994

190392490709135

308061521170129

498454011879264

806515533049393

1304969544928657

2111485077978050

3416454622906707

5527939700884757

8944394323791464

14472334024676221

23416728348467685

37889062373143906

61305790721611591

99194853094755497

160500643816367088

259695496911122585

420196140727489673

679891637638612258

1100087778366101931

1779979416004714189

2880067194370816120

4660046610375530309

7540113804746346429

12200160415121876738

19740274219868223167

31940434634990099905

51680708854858323072

83621143489848422977

135301852344706746049

218922995834555169026

354224848179261915075

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Answer 144 which is F(12) Reason 55 and 89 are the 10th and 11th Fibonacci numbers, If we add these we have 144 which is the 12 Fibonacci number and is a perfect square. I am using F(0) as the 0 Fibonacci number and F(1) as the first.

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The formula for the sum of an arithmetic sequence is ((first number) + (last number)) x (how many numbers) / 2, in this case, (1 + 100) x 100 / 2.The formula for the sum of an arithmetic sequence is ((first number) + (last number)) x (how many numbers) / 2, in this case, (1 + 100) x 100 / 2.The formula for the sum of an arithmetic sequence is ((first number) + (last number)) x (how many numbers) / 2, in this case, (1 + 100) x 100 / 2.The formula for the sum of an arithmetic sequence is ((first number) + (last number)) x (how many numbers) / 2, in this case, (1 + 100) x 100 / 2.

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The Fibonacci numbers below 100 are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987

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The first thing to notice is that all the numbers in the sequence are square numbers. 25=5x5 36=6x6 49=7x7 64=8x8 81=9x9 So the next three numbers to be squared are 10, 11 and 12. 10x10=100 11x11=121 12x12=144 Thus, the next three numbers in the sequence are 100, 121, 144 The equation for the sequence is (n+4)2

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S = 955 This is an arithmetic sequence, and the sum of an arithmetic sequence can be calculated as: S = n/2 x (U1 + Un) U1 is the first term (in this case 91) and Un is the last term (in this case 100). n presents the total number of terms in the sequence There are 10 numbers in this sequence (91, 92, 93, 94, 95, 96, 97, 98, 99, 100) So, the sum is : S = 10/2 x (91+100) = 955

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1, 2, 3, 5, 8, 13, 21, 34, 55, and 89 are.

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It is a sequence of numbers where the difference between successive numbers is decreasing by one.

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The sum of the first 100 cubed numbers is 25,502,500.

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The sum of the first 100 odd numbers is 10,000.

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The sum of the first 100 numbers, excluding zero, is 5,001.

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The sum of the first 100 positive even numbers is 10,100.

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The sum of the first 100 positive even numbers is 10,100.

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The sum of the cubes of the first 100 whole numbers is 25,502,500.

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The first numbers of pi is 3.14159265358979323846264338327950288419716939937510 58209749445923078164062862089986280348253421170679

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What is the 14th term in the arithmetic sequence in which the first is 100 and the common difference is -4? a14= a + 13d = 100 + 13(-4) = 48

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The first negative is -8. The first below -100 is -108.

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The sequence subtracts consecutive odd numbers. In this sequence it goes -1, -3, -5, -7. The next number will be 75 (84 - 9).

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The sequence is poorly defined. 1+3+5 appears to be a sequence of odd numbers. However, that cannot end in 100: it can attain the values of 99 or 101. Obviously the answer will depend on which one of these is the final number. An alternative is that the sequence is not that of odd numbers but some other sequence: for example, t(n) = (29n3 - 174n2 + 399n - 214)/40 which, for n = 1, 2, 3, generates the sequence 1, 3, 5, 11.35, 26.4, 54.5, 100 whose sum is 201

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This is very easy. Simply square the number: 100 squared (100 x 100) = 10,000. So, the sum of the first 100 odd numbers is 10,000.

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The sum of the first 100 counting numbers (1-100) is 5,001.

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The sum of the squares of the first 100 natural numbers [1..100] is 338350, while the sum of the first 100 natural numbers squared is 25502500.


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