<s>[INST] If $ \sum_{r=1}^{n} t_r $ = $ { n ( n + 1 ) ( n + 2 ) ( n + 3 ) } \over {8} $ , then what does $ \sum_{r=1}^{n} {1\over {t_r} } $ equal to? $ \sum_{r=1}^{n} t_r $ = $ { n ( n + 1 ) ( n + 2 ) ( n + 3 ) } \over {8} $ 
$ \sum_{r=1}^{n} {1\over {t_r} } $ = ? 
My progress so far : 
$ t_1 = {{1*2*3*4 } \over 8 } = 3 $ 
$ t_2 = 15 $ 
$ t_3 = 45 $ 
$ t_4 = 105 $ 
$ t_5 = 210 $ 
$ \sum_{r=1}^{n} {1\over {t_r} } $ = 1/3 + 1/12 + 1/30 + 1/60 + 1/105 .... So on , but I dont get a pattern . 
How do I solve this ? 
 [/INST] You can verify the series simplification at the bottom via telescoping or induction.
$$\sum_{r=1}^{n}t_r=\frac{n(n+1)(n+2)(n+3)}{8}$$
$$\implies t_n=\sum_{r=1}^{n}t_r-\sum_{r=1}^{n-1}t_r=\frac{n(n+1)(n+2)(n+3)}{8}-\frac{(n-1)n(n+1)(n+2)}{8}=\frac{n(n+1)(n+2)}{8}\big((n+3)-(n-1)\big)=\frac{n(n+1)(n+2)}{2}$$
$$\implies \frac{1}{t_r}=\frac{2}{r(r+1)(r+2)}=\frac{1}{r}-\frac{2}{r+1}+\frac{1}{r+2}$$
$$\implies \sum_{r=1}^n\frac{1}{t_r}=\sum_{r=1}^n\frac{1}{r}-2\sum_{r=1}^n\frac{1}{r+1}+\sum_{r=1}^n\frac{1}{r+2}=\frac{n(n+3)}{2(n+1)(n+2)}$$
</s>