二項式定理者，以究二項和之冪之拆解也。
( a + b ) n = ∑ k = 0 n [ C n k a n − k b k ] {\displaystyle (a+b)^{n}=\sum _{k=0}^{n}\left[C_{n}^{k}a^{n-k}b^{k}\right]}
可以數學歸納法證之：
n = 0 {\displaystyle n=0} ，則
( a + b ) 0 = 1 = ∑ k = 0 0 [ C 0 k a 0 − k b k ] {\displaystyle (a+b)^{0}=1=\sum _{k=0}^{0}\left[C_{0}^{k}a^{0-k}b^{k}\right]}
n = 1 {\displaystyle n=1} ，則
( a + b ) 1 = a + b = ∑ k = 0 1 [ C 1 k a 1 − k b k ] {\displaystyle (a+b)^{1}=a+b=\sum _{k=0}^{1}\left[C_{1}^{k}a^{1-k}b^{k}\right]}
令 n = j {\displaystyle n=j} ，則
( a + b ) j = ∑ k = 0 j [ C j k a j − k b k ] {\displaystyle (a+b)^{j}=\sum _{k=0}^{j}\left[C_{j}^{k}a^{j-k}b^{k}\right]}
n = j + 1 {\displaystyle n=j+1} ，則
( a + b ) j + 1 = ( a + b ) ∑ k = 0 j [ C j k a j − k b k ] {\displaystyle (a+b)^{j+1}=(a+b)\sum _{k=0}^{j}\left[C_{j}^{k}a^{j-k}b^{k}\right]}
= ∑ k = 0 j [ C j k a j + 1 − k b k ] + ∑ k = 0 j [ C j k a j − k b k + 1 ] {\displaystyle =\sum _{k=0}^{j}\left[C_{j}^{k}a^{j+1-k}b^{k}\right]+\sum _{k=0}^{j}\left[C_{j}^{k}a^{j-k}b^{k+1}\right]}
= a j + 1 + ∑ k = 1 j [ C j k a j + 1 − k b k ] + ∑ k = 1 j [ C j k − 1 a j + 1 − k b k ] {\displaystyle =a^{j+1}+\sum _{k=1}^{j}\left[C_{j}^{k}a^{j+1-k}b^{k}\right]+\sum _{k=1}^{j}\left[C_{j}^{k-1}a^{j+1-k}b^{k}\right]}
= a j + 1 + ∑ k = 1 j [ C j + 1 k a j + 1 − k b k ] {\displaystyle =a^{j+1}+\sum _{k=1}^{j}\left[C_{j+1}^{k}a^{j+1-k}b^{k}\right]}
= ∑ k = 0 j [ C j + 1 k a j + 1 − k b k ] {\displaystyle =\sum _{k=0}^{j}\left[C_{j+1}^{k}a^{j+1-k}b^{k}\right]}
故得證之。
![{\displaystyle (a+b)^{n}=\sum _{k=0}^{n}\left[C_{n}^{k}a^{n-k}b^{k}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd42e5f9515fa850ccbdc9510ed1eb88b041adc0)
，則
![{\displaystyle (a+b)^{0}=1=\sum _{k=0}^{0}\left[C_{0}^{k}a^{0-k}b^{k}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/54b5270efa23ebaf5c0fffd6d7ddb5f082784b22)
，則
![{\displaystyle (a+b)^{1}=a+b=\sum _{k=0}^{1}\left[C_{1}^{k}a^{1-k}b^{k}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/deb91faf425f0113145a170424a46d5bb2e037a1)
，則
![{\displaystyle (a+b)^{j}=\sum _{k=0}^{j}\left[C_{j}^{k}a^{j-k}b^{k}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff53dacf54caedf2daa3ee32c151dee70559e926)
，則
![{\displaystyle (a+b)^{j+1}=(a+b)\sum _{k=0}^{j}\left[C_{j}^{k}a^{j-k}b^{k}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dd4a7799e2ee5b90753b6dc3483439576b1f781a)
![{\displaystyle =\sum _{k=0}^{j}\left[C_{j}^{k}a^{j+1-k}b^{k}\right]+\sum _{k=0}^{j}\left[C_{j}^{k}a^{j-k}b^{k+1}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/625bee40623bc6437057e3513badbeed373f5527)
![{\displaystyle =a^{j+1}+\sum _{k=1}^{j}\left[C_{j}^{k}a^{j+1-k}b^{k}\right]+\sum _{k=1}^{j}\left[C_{j}^{k-1}a^{j+1-k}b^{k}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/819ddce7bd680a40a68bd32da55910c59686caa5)
![{\displaystyle =a^{j+1}+\sum _{k=1}^{j}\left[C_{j+1}^{k}a^{j+1-k}b^{k}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/229973af432645db60140d14121e535ee1a434ab)
![{\displaystyle =\sum _{k=0}^{j}\left[C_{j+1}^{k}a^{j+1-k}b^{k}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c5338ca8506c8c98af6deb64d795f150943b5451)
