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.
 A: 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
$$S_n=\frac {\left(2^n-1\right)^3 } {2^{3 n+1} }=\frac{1}{2} \left(1-2^{-n}\right)^3$$
A: Let's consider the sum
$$
  S(n) = \sum_{a = 1}^{n} \sum_{b = 1}^{n} \sum_{c = 1}^{n} \frac{ab(3a + c)}{2^a 2^b 2^c (a + b)(b + c)(c + a)}.
$$
The key is to look at what happens if we exchange the roles of $a$, $b$, and $c$. The sum is, for example, equal to
$$
  \sum_{a = 1}^{n} \sum_{c = 1}^{n} \sum_{b = 1}^{n} \frac{ac(3a + b)}{2^a 2^b 2^c (a + b)(b + c)(c + a)},
$$
which we obtain be replacing every $b$ with a $c$, and every $c$ with a $b$ in the original sum. By changing the order of summation, this implies that
$$
  S(n) = \sum_{a = 1}^{n} \sum_{b = 1}^{n} \sum_{c = 1}^{n} \frac{ac(3a + b)}{2^a 2^b 2^c (a + b)(b + c)(c + a)}.
$$
We do the same thing with all of the possible permutations of $a$, $b$, and $c$, and add the resulting expressions together. We get that
$$
  6S(n) = \sum_{a = 1}^{n} \sum_{b = 1}^{n} \sum_{c = 1}^{n} \frac{ab(3a + c) + ac(3a + b) + ab(3b + c) + bc(3b + a) + ac(3c + b) + bc(3c + a)}{2^a 2^b 2^c (a + b)(b + c)(c + a)}.
$$
At this point, a minor miracle occurs. It turns out that
$$
  ab(3a + c) + ac(3a + b) + ab(3b + c) + bc(3b + a) + ac(3c + b) + bc(3c + a)
$$
is equal to
$$
  3(a + b)(b + c)(c + a)
$$
and so we actually obtain the much simpler expression
$$
  6S(n) = \sum_{a = 1}^{n} \sum_{b = 1}^{n} \sum_{c = 1}^{n} \frac{3}{2^a 2^b 2^c} = 3 \sum_{a = 1}^{n} \frac{1}{2^a} \sum_{b = 1}^{n} \frac{1}{2^b} \sum_{c = 1}^{n} \frac{1}{2^c}
$$
and so
$$
  2S(n) = \left(\sum_{k = 1}^{n} \frac{1}{2^k} \right)^3 = \left(1 - \frac{1}{2^n} \right)^3
$$
and so finally
$$
  S(n) = \frac{1}{2} \left( 1 - \frac{1}{2^n} \right)^3
$$
as noticed by Claude Leibovici
