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

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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