using binomial theorem, the value of (0.999)³ correct to 3 decimal places is

$$({2^{3n}} - 1)$$ 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