If ∑r=1n tr=112n(n+1)(n+2), the value ∑r=1n 1tr is

If $\sum_{r = 1}^{n} t_{r} = \frac{1}{12} n (n + 1) (n + 2),$ the value $\sum_{r = 1}^{n} \frac{1}{t_{r}}$ is

A
$\frac{2 n}{n + 1}$
B
$\frac{n - 1}{(n + 1)!}$
C
$\frac{4 n}{n + 1}$
D
$\frac{3 n}{n + 2}$

Solution:

We have $t_{r} = \sum_{k = 1}^{r} t_{k} - \sum_{k = 1}^{r - 1} t_{k}$

$\begin{array}{l} = \frac{1}{12} r (r + 1) (r + 2) - \frac{1}{12} (r - 1) (r) (r + 1) \\ = \frac{1}{4} r (r + 1) \end{array}$

Now, $\frac{1}{t_{r}} = \frac{4}{r (r + 1)} = 4 (\frac{1}{r} - \frac{1}{r + 1})$

$\Rightarrow \sum_{r = 1}^{n} \frac{1}{t_{r}} = 4 \sum_{r = 1}^{n} (\frac{1}{r} - \frac{1}{r + 1}) = 4 (1 - \frac{1}{n + 1}) = \frac{4 n}{n + 1}$

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