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lfdahl
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Prove the binomial identity:
$$\sum_{j=0}^{n}(-1)^j{n \choose j}=0$$
- in two different ways
$$\sum_{j=0}^{n}(-1)^j{n \choose j}=0$$
- in two different ways
The only way I can think of is to do an induction proof. I haven't sat down to do it but it shouldn't be too hard.lfdahl said:Prove the binomial identity:
$$\sum_{j=0}^{n}(-1)^j{n \choose j}=0$$
- in two different ways
The binomial identity ∑(-1)^j(n choose j)=0 is a mathematical expression that represents the sum of alternating binomial coefficients, where n is a positive integer. It is also known as the Vandermonde's identity.
The binomial identity has many applications in combinatorics, probability, and algebra. It is often used to simplify complicated expressions and to prove other mathematical identities.
The binomial identity can be proved using mathematical induction or by using the properties of binomial coefficients. It can also be proved by using the binomial theorem or by using combinatorial arguments.
The (-1)^j term in the binomial identity alternates between positive and negative values as j increases. This helps to cancel out the terms in the sum, resulting in a final value of 0.
Yes, the binomial identity can be extended to any real or complex value of n using the generalized binomial theorem. However, for the identity ∑(-1)^j(n choose j)=0 to hold, n must be a positive integer.