chứng minh các hệ thức sau 

$C^{0}_{r}.C^{p}_{q} +C^{1}_{r}.C^{p - 1}_{q} + ... + C^{p}_{r}C^{0}_{q} = C^{p}_{r + q} $

$(C^{0}_{n})^{2} + (C^{1}_{n})^{2} +  ... + (C^{n}_{n})^{2} = C^{n}_{2n}$
 

Ta có:$(1+x)^{r}=C^{0}_{r}+C^{1}_{r}x+C^{2}_{r}x^{2}+...+C^{r}_{r}x^{r}$
         $(1+x)^{q}=C^{0}_{q}+C^{1}_{q}x+C^{2}_{q}x^{2}+...+C^{q}_{q}x^{q}$
Hệ số  $x^{p}$ trong tích $(1+x)^{r}(1+x)^{q}là:$
  $C^{0}_{r}.C^{p}_{q}+C^{1}_{0}.C^{p-1}_{q}+...+C^{p}_{r}.C^{0}_{q}$  (1)
ta lại có:$(1+x)^{r+p}=C^{0}_{r+q}+...+C^{p}_{r+q}x^{p}+...+C^{r+q}_{r+q}.x^{r+q}$
Hệ số $x^{p}$ trong $(1+x)^{r+q} $ là $C^{p}_{r+q}$  (2)
Mà $(1+x)^{r}.(1+x)^{q}=(1+x)^{q+r} nên từ $(1)$  va  $(2)$ \Rightarrow  dpcm$

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