The coefficient of $x^{49}$ in the expansion of $(x - 1)$$\left( {x\, - \,\frac{1}{2}\,} \right)$$\left( {x\, - \,\frac{1}{{{2^2}}}\,} \right)$ .....$\left( {x\, - \,\frac{1}{{{2^{49}}}}\,} \right)$ is equal to

If the expansion in powers of $x$ of the function  $\frac{1}{{\left( {1 - ax} \right)\left( {1 - bx} \right)}}$ is ${a_0} + {a_1}x + {a_2}{x^2} + \;{a_3}{x^3} + \; \ldots......$ then  ${a_n}$ is

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

${C_0} - {C_1} + {C_2} - {C_3} + ..... + {( - 1)^n}{C_n}$ is equal to

If  the number of terms in the expansion  of ${\left( {1 - \frac{2}{x} + \frac{4}{{{x^2}}}} \right)^n},x \ne 0$ is $28$ then the sum of the coefficients of all the terms in this expansion, is :

If ${C_0},{C_1},{C_2},.......,{C_n}$ are the binomial coefficients, then $2.{C_1} + {2^3}.{C_3} + {2^5}.{C_5} + ....$ equals