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In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial.According to the theorem, the power (+) expands into a polynomial with terms of the form , where the exponents and are nonnegative integers satisfying + = and the coefficient of each term is a specific positive integer ...
As with the (non-q) Chu–Vandermonde identity, there are several possible proofs of the q-Vandermonde identity.The following proof uses the q-binomial theorem.. One standard proof of the Chu–Vandermonde identity is to expand the product (+) (+) in two different ways.
In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question, and each with its own Boolean-valued outcome: success (with probability p) or failure (with probability q = 1 − p).
in which form it is clearly recognizable as an umbral variant of the binomial theorem (for more on umbral variants of the binomial theorem, see binomial type). The Chu–Vandermonde identity can also be seen to be a special case of Gauss's hypergeometric theorem, which states that
Download as PDF; Printable version; ... Fermat's identity or Chu's Theorem, [3] ... Further, by the binomial theorem, we also find that
Here, (+) is the binomial coefficient "p + 1 choose r", and the B j are the Bernoulli numbers with the convention that = +. The result: Faulhaber's formula [ edit ]
Since a binomial coefficient is always an integer, the nth binomial coefficient is divisible by p and hence equal to 0 in the ring. We are left with the zeroth and pth coefficients, which both equal 1, yielding the desired equation. Thus in characteristic p the freshman's dream is a valid identity.
Relationship to the binomial theorem [ edit ] The Leibniz rule bears a strong resemblance to the binomial theorem , and in fact the binomial theorem can be proven directly from the Leibniz rule by taking f ( x ) = e a x {\displaystyle f(x)=e^{ax}} and g ( x ) = e b x , {\displaystyle g(x)=e^{bx},} which gives