Factorial Sum

Let $S=\frac{3!+4!}{2(1!+2!)}+\frac{4!+5!}{3(2!+3!)}+\ldots+\frac{102!+103!}{101(100!+101!)}$ . Note that

$\frac{(n+2)!+(n+3)!}{(n+1)(n!+(n+1)!)}=\frac{(n+2)!(n+4)}{(n+1)n!(n+2)}=n+4.$

Therefore,

$S=\sum_{n=1}^{100}(n+4)=\boxed{5450}.$

if you observe the series than we can rewrite it as

$\frac { 3+3\times 4 }{ 3 }$ + $\frac { 4+4\times 5 }{ 5 }$ +.......... $\frac { 102+102\times 103 }{ 102 }$

therefore in general we can write it as $\frac { n+(n)(n+1) }{ n }$

$\therefore$ $\sum \limits^{n=102}_{n=3}\frac{n+n(n+1)}{n}$

$\therefore$ $\sum \limits^{n=102}_{n=3}(n+2)$

which is equal to 5250 + 200

= $\boxed { 5450 }$