Вычислите, буду благодарен. Created by stanmat. algebra-ru

Ряд арифметическая прогрессия, у которой .

Вычислите, буду благодарен...

igundane

$\dfrac{3!+4!}{2\, (1!+2!)}+\dfrac{4!+5!}{3\, (2!+3!)}+,,,+\dfrac{102!+103!}{101\, (100!+101!)}}=\\\\\\=5+6+7+...+104=\dfrac{5+104}{2}\cdot 100=54,5\cdot 100=5450\ ;\\\\\\\\a_{n}=\dfrac{(n+2)!+(n+3)!}{(n+1)\cdot (n!+(n+1)!)}=\dfrac{(n+2)!\cdot (1+(n+3))}{n!(n+1)+(n+1)(n+1)!}=\\\\\\=\dfrac{(n+2)!\cdot (n+4)}{(n+1)!+(n+1)(n+1)!}=\dfrac{(n+2)!\cdot (n+4)}{(n+1)!\cdot (1+n+1)}=\dfrac{(n+2)!\cdot (n+4)}{(n+1)!\cdot (n+2)}=\\\\\\=\dfrac{(n+2)!\cdot (n+4)}{(n+2)!}=n+4$