binomial coefficient formula
eq1:
$ \sum_{k=0}^n \binom{n}{k} = 2^n
proof
$ (1 + 1) ^ n = \sum_{k=0}^n \binom{n}{k}
$ (1 + 1) ^ n = 2^n
eq2:
$ \sum_{k=0}^n (-1)^k \binom{n}{k} = 0
proof
$ (-1 + 1) ^ n = \sum_{k=0}^n (-1)^k 1^{n-k} \binom{n}{k} = \sum_{k=0}^n (-1)^k \binom{n}{k}
$ (-1 + 1) ^ n = 0^n = 0
eq3:
$ \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k} = 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 \binom{n}{k} + \sum_{k=0}^n (-1)^k \binom{n}{k} = 2 \sum_{i=0}^{\lfloor n/2 \rfloor} \binom{n}{2i}
Since they cancel each other when k is odd
$ \sum_{i=0}^k \binom{n+i}{i} = \binom{n+k+1}{k}
eq5:
$ \sum_{i=0}^k \sum_{j=0}^l \binom{i+j}{i} = \binom{k+l+2}{k+1}-1
Pascal's triangulation
https://gyazo.com/161c01a67680a23524d71745a7cb5c38
$ \sum_{i=0}^k \binom{n}{i}\binom{m}{k - i} = \binom{n+m}{k}
special case(k=n, m=n)
$ \sum_{i=0}^n \binom{n}{i}^2 = \binom{2n}{n}
$ \sum_i \binom{i}{A}\binom{N-i-1}{B} = \binom{N}{A+B+1}
Ballistic Interpretation
https://gyazo.com/9aabb603ccaadea2194910347eb52fa4
eq6-3
$ \sum_i \binom{A + i}{i}\binom{B + K - i}{K - i} = \binom{A + B + K + 1}{K}
eq7:
$ \sum_{i=0}^k \binom{n+1}{i}\binom{m-i}{k - i} = \binom{n+m+1}{k}
$ \sum_i \binom{N-i}{i} = F_N
https://gyazo.com/68fc51e0aad6ed0f251979427ce9fbfe
eq8:
$ r\binom{n}{r} = n\binom{n-1}{r-1}
ref
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