Anyone ever research the summation of fibonacci sequence? Its properties are more amazing than 1,1,2,3,5, etc.
549823
Add one to everything and see what you get.\n\nYes, a *lot* of research has been done on the Fibonacci sequence. There is even a journal that was devoted to it (The Fibonacci Quarterly), although it has broadened to other number theory topics.
958618
is:http://en.wikipedia.org/wiki/Fibonacci_number\nclick to see.
1019687
F(n)=F(n-1)+F(n-2), where F(0) = 0 and F(1) = 1\nn= 2 \nF(2) = F(2-1) + F(2-2)\nF(2) = F(1) + F(0) \nF(2) = 1 + 0 = 1\nn = 3\nF(3) = F(3-1) + F(3-2)\nF(3) = F(2) + F(1)\nF(3) = 1 + 1 \nF(3) = 2\nn = 4\nF(4) = F(4-1) + F(4-2)\nF(4) = F(3) + F(2)\nF(4) = 2 + 1\nF(4) = 3\nn = 5\nF(5) = F(4) + F(3)\nF(5) = 3 + 2\nF(5) = 5
399866
There is an easy way to solve it. \n\nThese numbers are known as "pyramidal numbers" as each number in the sequence is equal to the number of blocks in an equilateral pyramid. As a side note, each sum in brackets is known as a "triangular number", as it is equal to the number of blocks in an equilateral triangle (note - a pyramid is a stack of triangles).\n\nMy memory is a little fuzzy on these things, but I believe that the closed-form equation is this:\n\nP(n) = n * (n + 1) * (n + 2) / 6
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This is the Gaussian binomial coefficient [n,1] for q=2. \n\nGreetings from The On-Line Encyclopedia of Integer Sequences!
881801
... 27, 26, 676, 678, 677, 458329, 458331, 458330, ...\n\nThis is sequence A059417 in the Online Encyclopedia of Integer Sequences.\n\nSquare, Add 2, Subtract 1, repeat...
385685
Consider the numbers\nA1,\nA1+A2,\nA1+A2+A3\n..\nA1+A2+A3+...+An.\n\nDo the computations modulo n. If none of these is divisible by n, then we have n numbers and n-1 non-zero residue classes (mod n), so there are two of these numbers which are equivalent (mod n) by the pigeon hole principle. But then subtracting gives a sum of the required form that is 0 (mod n), in other words, so that n divides that sum.\n\nLooking at the proof, there is no need for the sequence to be monotonic.
417496
It is divergant. Because the limit of the sentance to infinity is 1/2 and not zero. ( It should be zero so that the summation converges)
150456
here we are talking about a  sequence , so we try to get the formula of the sequence than the result.\nBut the proposed solution that 1=5 it's completly  wrong
619316
Do you have the same homework as Tilo B?\n\nStarting with 0, the sequence is the square of that number:\n\nf(n) = n^2. \n\nf(0) = 0 (0^2)\nf(1) = 1 (1^2)\nf(2) = 4 (2^2)\nf(3) = 9 (3^2)\nf(4) = 16 (4^2)\nf(5) = 25 (5^2)\netc.\n\nNote, it isn't clear if your indexes are going 0,1,2,3,4,... or are going in descending order 0,-1,-2,-3,-4... the resulting sequence would be the same.\n\nf(0) = 0 (0^2)\nf(-1) = 1 (-1^2)\nf(-2) = 4 (-2^2)\nf(-3) = 9 (-3^2)\nf(-4) = 16 (-4^2)\nf(-5) = 25 (-5^2)\netc.
19284
It's a notational convenience.\n\n1) It helps avoid the problem of running out of unique symbols when working with a large, possibly infinate, set of variables.\n\n2) It provides a way to keep track of ordering in a set of variables when that ordering makes a difference.\n\n3) It makes it possible to express the values of a sequence of variables when their value depends on their position in the sequence. An example would be the Fibonacci sequence where each number in the sequence is the sum of the two previous numbers (the first two being defined as 1). It can be expressed as: \n\na(sub)1 = 1\na(sub)2 = 1\na(sub)n = a(sub)(n-1) + a(sub)(n-2)
383876
In an arithmetic sequence, the formula for the nth term is\n\ninitial term + (common difference times (n-1))\n\nIn other words, you start with a number and keep adding the same number to that original number. In the sequence you gave (the odd numbers) the first number is 1 and you keeping adding 2. (The reason why the nth term is 1 + (n-1)2 is that you started adding 2 with the second term.)\n\nso the 101st term of that sequence would be\n\n1 + (100)2 = 201\n\nThe sum of the sequence, also known as an arithmetic series, has its  own formula:\n\nSum = (initial term + last term)*n/2\n\nThis comes out of the fact that if you add the first and last term, the result will be the same as adding the second and second-to-last term, which will be the same as adding the third and the third-to-last term, etc.\n\nSo for 20 terms of the sequence above, the first term is 1, the 20th term is 1+(2)19=39 and the sum of the sequence is (1+39)*20/2=400.\n\nI'll leave it up to you to do the sum of the first 30 terms.\n\nHope that helps.
174124
This geometric trick is very involving:\nYou can use it as a starting point for geometric construction of square roots, equivalence of areas and isometric transforms\nhttp://answers.yahoo.com/question/?qid=1006030501625\n\nThis other about divisibility and prime factoring of integers has deep mathematical concepts beyond it: starting from it you can speak about prime factoring, bynary count, combinatory calculus and mathematical induction\nhttp://answers.yahoo.com/question/?qid=1006022300298&r=w\n\nI'm going to use them in my school classes
849360
If you feel comfortable with geometry the golden ratio is good.  Fibonacci is good if you like sequences.  Pascal's triangle is good if you like factorials, binomials, and patterns.\n\nNote that the ration of consequtive fibonacci numbers converges to the golden ration as n-> infinity.\n\nHow about looking into how they are related?  I've shown you one.  I don't know how you'd relate Pascal's triangle to the other two, though.
517585
Starting with 0, the sequence is the square of that number:\n\nf(n) = n^2. \n\nf(0) = 0 (0^2)\nf(1) = 1 (1^2)\nf(2) = 4 (2^2)\nf(3) = 9 (3^2)\nf(4) = 16 (4^2)\nf(5) = 25 (5^2)\netc.\n\nNote, it isn't clear if your indexes are going 0,1,2,3,4,... or are going in descending order 0,-1,-2,-3,-4... the resulting sequence would be the same.\n\nf(0) = 0 (0^2)\nf(-1) = 1 (-1^2)\nf(-2) = 4 (-2^2)\nf(-3) = 9 (-3^2)\nf(-4) = 16 (-4^2)\nf(-5) = 25 (-5^2)\netc.
689300
Q98 = 1,1,2,3,5 \n\nP.S: Looks like the fibonacci series
49217
This is totally unfamiliar and I don't find it in my references.\n\nYou can prove that the series converges. \n\nn!/n^n = (1/n)(2/n)( n-2 other terms all < 1) < (1/n)(2/n)\nso an < 2/n^2. Since the sum(1/n^2) converges, your series converges. But that is as far as I am prepared to go.\n\nNote: If this were a Taylor series, since the nth term of the series is the nth derivative of f over n!, the nth derivative would have to have (n!)^2 in the numerator to cancel out the n! in the denominator.\n\nAs far as the answeres above, \na3 = (3*2*1) / (3^3) = 2/9, not 1/3.