binomial coefficient formula

eq1:
- $\sum_{k = 0}^{n} (k n) = 2^{n}$
- proof - binomial theorem
  - $(1 + 1)^{n} = \sum_{k = 0}^{n} (k n)$
  - $(1 + 1)^{n} = 2^{n}$
eq2:
- $\sum_{k = 0}^{n} (- 1)^{k} (k n) = 0$
- proof - binomial theorem
  - $(- 1 + 1)^{n} = \sum_{k = 0}^{n} (- 1)^{k} 1^{n - k} (k n) = \sum_{k = 0}^{n} (- 1)^{k} (k n)$
  - $(- 1 + 1)^{n} = 0^{n} = 0$
eq3:
- $\sum_{k = 0}^{⌊ n /2 ⌋} (2 k n) = 2^{n - 1}$
- proof
  - from eq1, eq2
  - $(1 + 1)^{n} + (- 1 + 1)^{n} = 2^{n}$
  - $(1 + 1)^{n} + (- 1 + 1)^{n} = \sum_{k = 0}^{n} (k n) + \sum_{k = 0}^{n} (- 1)^{k} (k n) = 2 \sum_{i = 0}^{⌊ n /2 ⌋} (2 i n)$
  - Since they cancel each other when k is odd
eq4: Field hockey stick identity
- $\sum_{i = 0}^{k} (i n + i) = (k n + k + 1)$
eq5:
- $\sum_{i = 0}^{k} \sum_{j = 0}^{l} (i i + j) = (k + 1 k + l + 2) - 1$
- Pascal’s triangulation
eq6: Vandermonde’s identity
- $\sum_{i = 0}^{k} (i n) (k - i m) = (k n + m)$
- special case(k=n, m=n)
  - $\sum_{i = 0}^{n} (i n)^{2} = (n 2 n)$
eq6-2 ARC110D
- $\sum_{i} (A i) (B N - i - 1) = (A + B + 1 N)$
- Ballistic Interpretation
eq6-3
- $\sum_{i} (i A + i) (K - i B + K - i) = (K A + B + K + 1)$
  - Attribute eq6-3 to eq6-2
  - Collapsing Duplicate Combinations
eq7:
- $\sum_{i = 0}^{k} (i n + 1) (k - i m - i) = (k n + m + 1)$
- eq7-proof
- Sum of binomial coefficients and Fibonacci
- $\sum_{i} (i N - i) = F_{N}$
eq8:
- $r (r n) = n (r - 1 n - 1)$
- https://mathtrain.jp/nikoukeisu

ref - A collection of counting techniques

Processing sums of binomial coefficients (formal power series) - Shino’s Blog

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binomial coefficient formula

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