Question

If $S_n={\sum }_{r=1}^n{T}_r=n(n+1)(n+2)(n+3)$, then ${\sum }_{r=1}^n\frac{1}{T}$ is equal to:

Answer 1

Let $T_n=4n(n+1)(n+2)T_n=4n(n+1)(n+2)\frac{(n+3)-(n+1)}{4}{T}_n=n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)T_1=1\cdot 2\cdot 3\cdot 4-0T_2=2\cdot 3\cdot 4\cdot 5-1\cdot 2\cdot 3\cdot 4T_3=3\cdot 4\cdot 5\cdot 6-2\cdot 3\cdot 4\cdot 5T_n=n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)$

Adding all, we get

$S_n=n(n+1)(n+2)(n+3)$

$\frac{1}{T_k}=\frac{1}{4k(k+1)(k+2)}=\frac{1}{4}[\frac{1}{2}[\frac{1}{k+2}-\frac{1}{k+1}]-\frac{1}{2}[\frac{1}{k+1}-\frac{1}{k}]]\frac{1}{T_k}=\frac{1}{8}[[\frac{1}{k+2}-\frac{1}{k+1}]-[\frac{1}{k+1}-\frac{1}{k}]]\frac{1}{T_1}=\frac{1}{8}[[\frac{1}{3}-\frac{1}{2}]-[\frac{1}{2}-\frac{1}{1}]]\frac{1}{T_2}=\frac{1}{8}[[\frac{1}{4}-\frac{1}{3}]-[\frac{1}{3}-\frac{1}{2}]]\frac{1}{T_n}=\frac{1}{8}[[\frac{1}{n+2}-\frac{1}{n+1}]-[\frac{1}{n+1}-\frac{1}{n}]]{\sum }_{k=1}^n\frac{1}{T_k}=\frac{1}{8}[[\frac{1}{n+2}-\frac{1}{2}]-[\frac{1}{n+1}-1]]{\sum }_{k=1}^n\frac{1}{T_k}=\frac{1}{8}[\frac{1}{n+2}-\frac{1}{n+1}+\frac{1}{2}]{\sum }_{k=1}^n\frac{1}{T_k}=\frac{-2+(n+2)(n+1)}{16(n+1)(n+2)}=\frac{n^2+3n}{16(n+1)(n+2)}$