Symbol newtona - rozwiąż działania

[knzxo]

n!+(n+2)!
(n+1)!-(n-1)!


(2k)-(2k-1)!
(2k+2)!-(2k+1)!

[panb]

\(\ldots=\frac{n!(1+(n+1)(n+2))}{(n-1)!(n(n+1)-1)}= \frac{n+n(n+1)(n+2)}{n(n+1)+1}= \frac{n(n^2+3n+3)}{n^2+n+1}\)

[eresh]

(2k)-(2k-1)!
(2k+2)!-(2k+1)!

\(\frac{(2k)!-(2k-1)!}{(2k+2)!-(2k+1)!}=\frac{(2k-1)!2k-(2k-1)!}{(2k+1)!(2k+2)-(2k+1)!}=\frac{(2k-1)!(2k-1)}{(2k+1)!(2k+2-1)}=\frac{(2k-1)!(2k-1)}{(2k-1)!(2k)(2k+1)(2k+1)}=\\=\frac{2k-1}{2k(2k+1)^2}\)