\(2^{2016}=\left(2^{6}\right)^{336}=(64)^{336}=(63+1)^{336}\)
\((63+1)^{336}=^{366}C_0(63)^0(1)^{366}+^{366}C_1(63)^1(1)^{364}+^{366}C_2(63)^2(1)^{362}+…… = 1+63K\)
\(\therefore\) Remainder \(=1\)
So, the correct option is (A): \(1\)
\[ \left( \frac{1}{{}^{15}C_0} + \frac{1}{{}^{15}C_1} \right) \left( \frac{1}{{}^{15}C_1} + \frac{1}{{}^{15}C_2} \right) \cdots \left( \frac{1}{{}^{15}C_{12}} + \frac{1}{{}^{15}C_{13}} \right) = \frac{\alpha^{13}}{{}^{14}C_0 \, {}^{14}C_1 \cdots {}^{14}C_{12}} \]
Then \[ 30\alpha = \underline{\hspace{1cm}} \]
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
