Evaluating $\sum_{a=1}^6\sum_{b=1}^6\sum_{c=1}^6\frac{ab(3a+c)}{2^a2^b2^c(a+b)(b+c)(c+a)}$ without a calculator? Is there a way to get this value without calculator?
$$\sum_{a=1}^6\sum_{b=1}^6\sum_{c=1}^6\frac{ab(3a+c)}{2^a2^b2^c(a+b)(b+c)(c+a)}$$
I'm currently studying AIME.

Consider
$$S_n=\sum_{a=1}^n\sum_{b=1}^n\sum_{c=1}^n\frac{ab(3a+c)}{2^a\,2^b\,2^c\,(a+b)(b+c)(c+a)}$$ and compute $S_n$ for the very first values of $n$. This gives the sequence
$$\left\{\frac{1}{16},\frac{27}{128},\frac{343}{1024},\frac{3375}{8192}\right\}$$ and you can notice that the numerators are cubes and that the denominators are powers of $2$.
So, you can conjecture that
$$S_n=\frac {\left(2^n-1\right)^3 } {2^{3 n+1} }=\frac{1}{2} \left(1-2^{-n}\right)^3$$
