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  2. 1/2 + 1/4 + 1/8 + 1/16 + ⋯ - Wikipedia

    en.wikipedia.org/wiki/1/2_%2B_1/4_%2B_1/8_%2B_1/...

    1/2 + 1/4 + 1/8 + 1/16 + ⋯. First six summands drawn as portions of a square. The geometric series on the real line. In mathematics, the infinite series ⁠ 1 2 ⁠ + ⁠ 1 4 ⁠ + ⁠ 1 8 ⁠ + ⁠ 1 16 ⁠ + ··· is an elementary example of a geometric series that converges absolutely. The sum of the series is 1. In summation notation ...

  3. Central binomial coefficient - Wikipedia

    en.wikipedia.org/wiki/Central_binomial_coefficient

    The central binomial coefficients give the number of possible number of assignments of n -a-side sports teams from 2 n players, taking into account the playing area side. The central binomial coefficient is the number of arrangements where there are an equal number of two types of objects. For example, when , the binomial coefficient is equal ...

  4. Catalan number - Wikipedia

    en.wikipedia.org/wiki/Catalan_number

    The Catalan numbers can be interpreted as a special case of the Bertrand's ballot theorem. Specifically, is the number of ways for a candidate A with n+1 votes to lead candidate B with n votes. The two-parameter sequence of non-negative integers is a generalization of the Catalan numbers.

  5. Binomial theorem - Wikipedia

    en.wikipedia.org/wiki/Binomial_theorem

    In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial.According to the theorem, it is possible to expand the polynomial (x + y) n into a sum involving terms of the form ax b y c, where the exponents b and c are nonnegative integers with b + c = n, and the coefficient a of each term is a specific positive integer depending ...

  6. Bertrand's postulate - Wikipedia

    en.wikipedia.org/wiki/Bertrand's_postulate

    Bertrand's postulate was proposed for applications to permutation groups. Sylvester (1814–1897) generalized the weaker statement with the statement: the product of k consecutive integers greater than k is divisible by a prime greater than k. Bertrand's (weaker) postulate follows from this by taking k = n, and considering the k numbers n + 1 ...

  7. Proof of Bertrand's postulate - Wikipedia

    en.wikipedia.org/wiki/Proof_of_Bertrand's_postulate

    Assume that there is a counterexample: an integer n2 such that there is no prime p with n < p < 2n. If 2n < 427, then p can be chosen from among the prime numbers 3, 5, 7, 13, 23, 43, 83, 163, 317, 631 (each being the largest prime less than twice its predecessor) such that n < p < 2n. Therefore, n ≥ 427.

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