Ifn frac12 * 3^10 + frac12^2 * 3^9 + ... .. + frac12^10 * 3 = fracK2^10 * 3^10,n then the remainder when K is divided by 6 is

n n frac12 * 3^10+(1 / 2^2 * 3^9)+(1 / 2^3 * 3^8)+ ... ... +frac12^10 * 3=frac K2^10 * 3^10 n K=2^9+2^8 * 3+2^7 * 3^2+ ... ... +3^9 n =frac2^9(((3 / 2))^10-1)(3 / 2)-1=3^10-2^10n nn nNow, 3^10-2^10=(3^5-2^5)(3^5+2^5)nn n =(211)(275) n =(35 x 6+1)(45 x 6+5) n =6 lambda+5n n

If \frac{1}{2 \cdot 3^{10}} + \frac{1}{2^{2} \cdot 3^... | Sequence & Series (Mathematics)

\n\begin{aligned}\n& \frac{1}{2 \cdot 3^{10}}+\frac{1}{2^2 \cdot 3^9}+\frac{1}{2^3 \cdot 3^8}+\ldots \ldots+\frac{1}{2^{10} \cdot 3}=\frac{\mathrm{K}}{2^{10} \cdot 3^{10}} \\\n& K=2^9+2^8 \cdot 3+2^7 \cdot 3^2+\ldots \ldots+3^9 \\\n& =\frac{2^9\left(\left(\frac{3}{2}\right)^{10}-1\right)}{\frac{3}{2}-1}=3^{10}-2^{10}\n\end{aligned}\n

\n \nNow,

3^{10}-2^{10}=\left(3^5-2^5\right)\left(3^5+2^5\right)

\n\begin{aligned}\n& =(211)(275) \\\n& =(35 \times 6+1)(45 \times 6+5) \\\n& =6 \lambda+5\n\end{aligned}\n

Explanation

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