[tex]\it \Big C_n^n +\Big C_n^{n-1}+\Big C_n^{n-2}=\Big C_n^{n-n}+\Big C_n^{n-n+1}+\Big C_n^{n-n+2}=\Big C_n^0+\Big C_n^1+\Big C_n^2=\\ \\ \\ =1+n+\dfrac{n!}{2!(n-2)!}=1+n+\dfrac{(n-2)!(n-1)n}{2(n-2)!}=1+n+\dfrac{(n-1)n}{2}[/tex]
Acum, ecuația devine:
[tex]\it 1+n+\dfrac{(n-1)n}{2}=22|_{-1} \Rightarrow n+\dfrac{(n-1)n}{2}=21|_{\cdot2} \Rightarrow 2n+(n-1)n=42 \Rightarrow \\ \\ \\ \Rightarrow n(2+n-1)=42\Rightarrow n(n+1)=42=6\cdot7\Rightarrow n=6[/tex]
