Playing with 2

$\begin{aligned} S & = \sum_{n=-\infty}^6 2^n \\ & = \sum_{n=-\infty}^0 2^n + \sum_{n=1}^6 2^n \\ & = \sum_{n=0}^\infty 2^{-n} + 2 \sum_{n=0}^5 2^n \\ & = \sum_{n=0}^\infty \left(\frac 12\right)^n + 2 \left(\frac {2^6-1}{2-1}\right) \\ & = \frac 1{1-\frac 12} + 126 \\ & = 2+126 = 128 = 2^7 \\ & = 3^{\log_3 2^7} \end{aligned}$

Therefore, $abc-1=3(2)(7)-1 = \boxed{41}$ .

We know that

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

which

$\displaystyle \dfrac{1}{2^{n}} = 2^{-n}$

and

$\displaystyle \sum_{n=0}^{N} 2^{n}$ = $2^{N+1} - 1$

Then, add up these sequence

$\displaystyle \sum_{n=-{\infty}}^{N} 2^{n} = 2^{N+1}$

putting n with 6 and we get $\displaystyle \sum_{n=-{\infty}}^{6} = 2^{7}$

and $2^{7} = 3^{s}$

putting $log$ on both side

we get $\dfrac {\log{2^{7}}}{\log{3}} = s$

which is equal to

$\log_{3}{2^{7}} = s$ so we got $abc$ is $42$ and $abc - 1$ is $\displaystyle \boxed {41}$