\[\begin{cases} \displaystyle \sum_{i=1}^{\infty}\frac{i^{2}}{2^{i}}=m \\ \displaystyle\frac{\displaystyle \sum_{i = 1}^{m^2}i}{\displaystyle \sum_{i = 1}^{m^2}i^3} = \frac{R}{Z} \end{cases}\]

Here \(R\) and \(Z\) are positive co-prime integers.

Find \(m^{2}+R+Z\).

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