$$({2^{3n}} - 1)$$ will be divisible by $$(\forall n \in N)$$

If in the expansion (a $$-$$ 2b)ⁿ, the sum of the 5^th and 6^th term is zero, then the value of $${a \over b}$$ is

Sum of the last 30 coefficients in the expansion of (1 + x)⁵⁹, when expanded in ascending powers of x is

If $${(1 - x + {x^2})^n} = {a_0} + {a_1}x + {a_2}{x^2} + \,\,....\,\,{a_{2n}}{x^{2n}}$$, then the value of $${a_0} + {a_2} + {a_4} + \,\,....\,\,{a_{2n}}$$ is