Question

Care sunt răspunsurile la exercițiul 3 complet și 4 punctul a???

Answer 1

$\frac{(n+1)!}{(n-2)!}= \frac{(n-2)!(n-1)n(n+1)}{(n-1)!}=(n-1)n(n+1)=n(n^2-1)=n^3-1 \\ \frac{(n-2)!}{(n-5)!} = \frac{(n-5)!(n-4)(n-3)(n-2)}{(n-5)!}=(n-4)(n-3)(n-2)= \\ =(n^2-7n+12)(n-2)=n^3-9n^2-2n-24 \\ \frac{1}{n!}- \frac{2}{(n+1)!} + \frac{1}{(n+2)!} = \frac{(n+1)(n+2)-2(n+2)+1}{n!(n+1)(n+2)}= \frac{n^2+3n+2-2n-4+1}{(n+2)!} = \\ = \frac{n^2+n+1}{(n+2)!} \\ ex.4 \\ \frac{7!4!}{10!} ( \frac{8!}{3!5!}- \frac{9!}{2!7!} })= \frac{7!4!}{7!8*9*10}( \frac{5!**6*7*8}{5!*6}- \frac{7!*8*9}{7!*2})=$
$= \frac{1}{30}(56-36)= \frac{20}{30}= \frac{2}{3} \\ \frac{5!}{m(m+1)}* \frac{(m+1)!}{(m-1)!*3!}= \frac{120}{m(m+1)}* \frac{(m-1)!m(m+1)}{(m-1)!*6}=20$