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Q)

If $a_{n}=\displaystyle\sum_{k=1}^{n}\;\large\frac{1}{k(n+1-k)}$, then for $n\;\geq\;2$

$(a)\;a_{n+1} > a_{n}\qquad(b)\;a_{n+1} = a_{n}\qquad(c)\;a_{n+1} < a_{n}\qquad(d)\;a_{n+1} - a_{n}=n$

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