Another Summation (Part 2)

Algebra Level 4

m=1m22m \large \displaystyle \sum_{m=1}^{\infty} \dfrac{m^2}{2^m}

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

P.S. Don't get tricked!

Try another problem on my set Let's Practice

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