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

Answer: (D)

5

Answer 2

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

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

= $\frac {3^{10}−2^{10}}{2^{10}⋅3^{10}} = \frac {K}{2^{10}⋅3^{10}}$
$K = 3^{10}−2^{10}$
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