#A08 · Algebra & Number Theory · Lv.2View in 3D Mind-Map

Binomial Theorem & Pascal's Triangle

Blaise Pascal & Isaac Newton · 1654 CE

(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

\binom{n}{k} = \frac{n!}{k!(n-k)!} = \binom{n-1}{k-1} + \binom{n-1}{k}

Expands any power of a binomial (a+b)^n using combinatorial coefficients, bridging algebra, combinatorics, binomial probability, and calculus.

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Power n

Term index k

\binom{5}{2} = \frac{5!}{2!(5-2)!} = 10, \quad \sum_{j=0}^{5}\binom{5}{j} = 2^{5} = 32

이항계수 C(5,2)

파스칼 행 총합 2^5

대칭 항 비중

31.25%

Prerequisite Path (4 STEP)

Click steps in order to follow the 3D neon path from basics to target.

ax + b = 0 \implies x = -\frac{b}{a}

(a+b)^2 = a^2 + 2ab + b^2

a^m \cdot a^n = a^{m+n}

(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

n개의 (a+b) 괄호 중에서 b를 k번 선택하고 나머지 (n-k)번은 a를 선택하는 경우의 수가 바로 조합 nCk 이므로, a^(n-k)b^k 항의 계수는 정확히 nCk 가 됩니다.

Bernstein basis polynomials for 3D Bézier curves and ensemble classifier majority voting

#Bernstein Basis#Dynamic Programming#Combinatorial AI

const nCr = (n: number, k: number): number => k === 0 ? 1 : (n * nCr(n - 1, k - 1)) / k;