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Mathematics Question on Algebra of Complex Numbers

If 6312+10311+20310+4039+⋯+102403=2n⋅m,\begin{array}{l} \frac{6}{3^{12}}+\frac{10}{3^{11}}+\frac{20}{3^{10}}+\frac{40}{3^9}+\cdots +\frac{10240}{3}=2^n\cdot m, \end{array}where m is odd, then m ⋅ n is equal to ________.

Answer

1312+5(20312+21311+22310+⋯+2113)=2n⋅m\begin{array}{l} \frac{1}{3^{12}}+5\left(\frac{2^0}{3^{12}}+\frac{2^1}{3^{11}}+\frac{2^2}{3^{10}}+\cdots +\frac{2^{11}}{3}\right)=2^n\cdot m\end{array}

⇒ 1312+5(1312((6)2−1)(6−1))=2n⋅m\begin{array}{l} \Rightarrow\ \frac{1}{3^{12}}+5\left(\frac{1}{3^{12}}\frac{\left(\left(6\right)^2-1\right)}{\left(6-1\right)}\right)=2^n\cdot m\end{array}

⇒ 1312+55(1312⋅212⋅312−1312)=2n⋅m\begin{array}{l} \Rightarrow\ \frac{1}{3^{12}}+\frac{5}{5}\left(\frac{1}{3^{12}}\cdot2^{12}\cdot 3^{12}-\frac{1}{3^{12}}\right)=2^n\cdot m\end{array}

⇒ 1312+212−1312=2n⋅m\begin{array}{l} \Rightarrow\ \frac{1}{3^{12}}+2^{12}-\frac{1}{3^{12}}=2^n\cdot m\end{array}

⇒ 1312+212−1312=2n⋅m\begin{array}{l} \Rightarrow\ \frac{1}{3^{12}}+2^{12}-\frac{1}{3^{12}}=2^n\cdot m\end{array}

⇒ 2 n ⋅ m = 212

⇒ m = 1 and n = 12

m ⋅ n = 12