\[\large \displaystyle \sum_{n=1}^{2015} \dfrac{1}{T_{n}}\]

If \(T_{n} = \frac{1}{4} (n+2)(n+3)\), the value of the above sum is in the form \(\dfrac{A}{B}\), where \(A,B \in \mathbb{N}\) and \(\gcd(A,B)=1\). Find \(A+B\).

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