If \( \sum_{r=1}^{n} t_{r}=\frac{n(n+1)(n+2)(n+3)}{8} \), then \( \...

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If \( \sum_{r=1}^{n} t_{r}=\frac{n(n+1)(n+2)(n+3)}{8} \), then \( \sum_{r=1}^{n} \frac{1}{t_{r}} \) equals
\( \mathrm{P} \)
(A) \( -\left(\frac{1}{(n+1)(n+2)}-\frac{1}{2}\right) \)
(B) \( \left(\frac{1}{(n+1)(n+2)}-\frac{1}{2}\right) \)
(C) \( \left(\frac{1}{(n+1)(n+2)}+\frac{1}{2}\right) \)
(D) \( \left(\frac{1}{(n-1)(n-2)}+\frac{1}{2}\right) \)
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