\[ \large \displaystyle \sum_{m=1}^{\infty} \dfrac{m^2}{2^m} \]

The summation above is equal to \( \dfrac{a}{b} + c\), where \(a,b,c\) are integers, with \(a\) and \(b\) positive and \( \text{gcd}(a,b) = 1 \), and \( 0 \leq c <1 \). Find \(a+b-c\).

**P.S.** Don't get tricked!

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