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2 Answers

S=1/2+1/4+1/8+...+1/8192.

S/2=1/4+1/8+1/16+...+1/8192+1/16384.

S-S/2=S/2=1/2-1/16384=8192/16384-1/16384=8191/16384.

So S=2×8191/16384=8191/8192.
by Top Rated User (679k points)

Let the given sum be $S$. By inspection, we find that $2^6 \equiv 64 \equiv -1 \pmod{13}$, so $2^{-6} \equiv (-1)^{-1} \equiv -1 \pmod{13}$. It follows that $2^{-5} \equiv 2 \cdot 2^{-6} \equiv 2 \cdot -1 \equiv -2 \pmod{13}$, and that $2^{-4} \equiv -4 \pmod{13}$, and so forth. Thus,$$S \equiv -2^5 - 2^4 - 2^3 - 2^2 - 2 - 1 \equiv -63 \equiv \boxed{2} \pmod{13}$$

Solution 2:

Let the given sum be $S$. The derivation of the geometric series formula remains the same if we consider the result $\mod{13}$. In particular,\begin{align*}S &\equiv 2^{-1} \cdot (1 + 2^{-1} + \cdots + 2^{-5}) \pmod{13} \\ 2^{-1} S &\equiv 2^{-1} \cdot (2^{-1} + \cdots + 2^{-5} + 2^{-6}) \pmod{13}.\end{align*}Subtracting the two yields that$$(1 - 2^{-1}) \cdot S \equiv 2^{-1} \cdot (1 - 2^{-6}) \pmod{13}.$$Since $2 \cdot (1-2^{-1}) \equiv 2 - 1 \equiv 1 \pmod{13}$, it follows that $1-2^{-1} \equiv 2^{-1} \pmod{13}$, so$$S \equiv 2^{-1} \cdot \frac{1 - 2^{-6}}{1-2^{-1}} \equiv 1 - 2^{-6} \equiv \frac{63}{64} \equiv \frac{-2}{-1} \equiv \boxed{2} \pmod{13}.$$*AoPS answer

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