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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

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Correct option is C

We have tr=∑k=1r tk−∑k=1r−1 tk=112r(r+1)(r+2)−112(r−1)(r)(r+1)=14r(r+1)Now, 1tr=4r(r+1)=41r−1r+1⇒ ∑r=1n 1tr=4∑r=1n 1r−1r+1=41−1n+1=4nn+1

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

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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