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  • If <img alt="\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}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7B

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:

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

5


Answers (1)

best_answer

\mathrm{ \frac{1}{2 \cdot 3^{10}}+\frac{1}{2^{2} \cdot 3^{9}}+\cdots+\frac{1}{2^{9} \cdot 3^{2}}+\frac{1}{2^{10} \cdot 3}}

This is a GP with common ration = 3/2 and n = 10

So, Sum =\frac{3^{10}-2^{10}}{2^{10} \cdot 3^{10}}

\mathrm{ So, k=3^{10}-2^{10} }

Since we need the remainder when \mathrm{k}  is divided by 6

\mathrm{So,\: 3^{10}=6 q_{1}+3 \: \: and\: \: 2^{10}=6 q_{2}+4}

Now \mathrm{k} will be of the form \mathrm{\left(6 q_{1}+3\right)-\left(6 q_{2}+4\right)}
\mathrm{=6\left(q_{1}-q_{2}\right)-1}
Hence, when \mathrm{k} is divided by 6 , we get the remainder as \mathrm{6-1=5}

Posted by

Suraj Bhandari

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