Chứng minh công thức nhị thức Newton bằng phương pháp quy nạp:

Thật vậy, theo giả thiết quy nạp ta có:

${(a+b)}^k=C_k^0{a}^k+C_k^1{a}^{k-1}b+\mathrm{...}+C_k^{k-1}a{b}^{k-1}+C_k^k{b}^k.$

Khi đó:

${(a+b)}^{k+1}=\left(a+b\right){\left(a+b\right)}^k$

$=a{\left(a+b\right)}^k+b{\left(a+b\right)}^k$

$=a\left(C_k^0{a}^k+C_k^1{a}^{k-1}b+\mathrm{...}+C_k^{k-1}a{b}^{k-1}+C_k^k{b}^k\right)$

$+b\left(C_k^0{a}^k+C_k^1{a}^{k-1}b+\mathrm{...}+C_k^{k-1}a{b}^{k-1}+C_k^k{b}^k\right)$

$=\left(C_k^0{a}^{k+1}+C_k^1{a}^{k}b+C_k^2{a}^{k-1}{b}^2+\mathrm{...}+C_k^{k-1}{a}^2{b}^{k-1}+C_k^{k}a{b}^k\right)$

$+\left(C_k^0{a}^{k}b+C_k^1{a}^{k-1}{b}^2+\mathrm{...}+C_k^{k-2}{a}^2{b}^{k-1}+C_k^{k-1}a{b}^k+C_k^k{b}^{k+1}\right)$

$=C_k^0{a}^{k+1}+\left(C_k^0+C_k^1\right)a^{k}b+\left(C_k^1+C_k^2\right)a^{k-1}{b}^2+\mathrm{...}$

$+\left(C_k^{k-2}+C_k^{k-1}\right)a^2{b}^{k-1}+\left(C_k^{k-1}+C_k^k\right)a{b}^k+C_k^k{b}^{k+1}$

$=1.a^{k+1}+C_{k+1}^1{a}^{k}b+C_{k+1}^2{a}^{k-1}{b}^2+\mathrm{...}+C_{k+1}^{k-1}{a}^2{b}^{k-1}+C_{k+1}^{k}a{b}^k+1.b^{k+1}$

$(vì C_k^i+C_k^{i+1}=C_{k+1}^{i+1}\forall 0\le i\le k, i\in \mathrm{\mathbb{N}}, k\in \mathrm{\mathbb{N}}*)$

$=C_{k+1}^0{a}^{k+1}+C_{k+1}^1{a}^{(k+1)-1}b+\mathrm{...}+C_{k+1}^{(k+1)-1}a{b}^{(k+1)-1}+C_{k+1}^{k+1}{b}^{k+1}.$