If frac{1}{2 cdot 3^{10}}+frac{1}{2^{2} cdot | vedclass

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(D) The given expression is a geometric series: $\sum_{n=1}^{10} \frac{1}{2^n \cdot 3^{11-n}} = \frac{K}{2^{10} \cdot 3^{10}}$.Multiplying both sides

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$5$

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(D) The given expression is a geometric series: $\sum_{n=1}^{10} \frac{1}{2^n \cdot 3^{11-n}} = \frac{K}{2^{10} \cdot 3^{10}}$.Multiplying both sides by $2^{10} \cdot 3^{10}$,we get $K = \sum_{n=1}^{10} 2^{10-n} \cdot 3^{n-1} = 3^0 \cdot 2^9 + 3^1 \cdot 2^8 + \ldots + 3^9 \cdot 2^0$.This is a geometric progression with first term $a = 2^9$,common ratio $r = \frac{3}{2}$,and $n = 10$ terms.$K = \frac{2^9 ((\frac{3}{2})^{10} - 1)}{\frac{3}{2} - 1} = \frac{2^9 (\frac{3^{10}}{2^{10}} - 1)}{\frac{1}{2}} = 2^{10} \cdot \frac{3^{10} - 2^{10}}{2^{10}} = 3^{10} - 2^{10}$.We need to find the remainder when $K = 3^{10} - 2^{10}$ is divided by $6$.$K = 3^{10} - 2^{10} = (3^5 - 2^5)(3^5 + 2^5) = (243 - 32)(243 + 32) = (211)(275)$.$211 = 6 	imes 35 + 1$,so $211 \equiv 1 \pmod{6}$.$275 = 6 	imes 45 + 5$,so $275 \equiv 5 \pmod{6}$.$K \equiv 1 	imes 5 \equiv 5 \pmod{6}$.Thus,the remainder is $5$.

If $\frac{1}{2 \cdot 3^{10}}+\frac{1}{2^{2} \cdot 3^{9}}+\ldots+\frac{1}{2^{10} \cdot 3}=\frac{K}{2^{10} \cdot 3^{10}}$,then the remainder when $K$ is divided by $6$ is

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