Question

If \(\frac {1}{2⋅3^{10}}+\frac {1}{2^2⋅3^9}+⋯+\frac {1}{2^{10}⋅3} = \frac {K}{2^{10}⋅3^{10}}.\) Then the remainder when \(K\) is divided by \(6\) is:

• A. 1
• B. 2
• C. 3
• D. 5

Answer 1

The Correct Option is D

\(\frac {1}{2⋅3^{10}}+\frac {1}{2^2⋅3^9}+⋯+\frac {1}{2^{10}⋅3} = \frac {K}{2^{10}⋅3^{10}}\)

\(\frac {1}{2⋅3^{10}}[\frac {(\frac {3}{2}^{10})−1}{\frac 32−1}] =  \frac {K}{2^{10}⋅3^{10}}\)

= \(\frac {3^{10}−2^{10}}{2^{10}⋅3^{10}} = \frac {K}{2^{10}⋅3^{10}}\)
\(K = 3^{10}−2^{10}\)
Now,
K = (1 + 2)¹⁰ – 2¹⁰
K = ¹⁰C₀ + ¹⁰C₁ 2 + ¹⁰C₂ 2³ + … + ¹⁰C₁₀ 2¹⁰ – 2¹⁰
K = ¹⁰C₀ + ¹⁰C₁ 2 + 6λ + ¹⁰C₉⋅ 2⁹
K = 1 + 20 + 5120 + 6λ
K = 5136 + 6λ + 5
K = 6μ + 5
Where λ, μ ∈ N
∴ remainder = 5

So, the correct option is (D): 5