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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 ...
The hockey stick identity confirms, for example: for n=6, r=2: 1+3+6+10+15=35. In combinatorics , the hockey-stick identity , [ 1 ] Christmas stocking identity , [ 2 ] boomerang identity , Fermat's identity or Chu's Theorem , [ 3 ] states that if n ≥ r ≥ 0 {\displaystyle n\geq r\geq 0} are integers, then
In mathematics, Pascal's rule (or Pascal's formula) is a combinatorial identity about binomial coefficients.It states that for positive natural numbers n and k, + = (), where () is a binomial coefficient; one interpretation of the coefficient of the x k term in the expansion of (1 + x) n.
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.
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 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
A binomial raised to the n th power, represented as (x + y) n can be expanded by means of the binomial theorem or, equivalently, using Pascal's triangle. For example, the square (x + y) 2 of the binomial (x + y) is equal to the sum of the squares of the two terms and twice the product of the terms, that is:
This is because if a and b are nilpotent elements of R with a n = 0 and b m = 0, and r is any element of R, then (a·r) n = a n ·r n = 0, and by the binomial theorem, (a+b) m+n = 0. Therefore, the set of all nilpotent elements forms an ideal known as the nil radical of a ring.