Let $\left(\frac{n}{k}\right)=\frac{n !}{k !(n-k) !}$. Then the sum $\frac{1}{2^{10}} \sum \limits_{ k =0}^{10}\left(\frac{10}{ k }\right) k ^2$, lies in the interval

If $\sum_{r=1}^{10} r !\left( r ^{3}+6 r ^{2}+2 r +5\right)=\alpha(11 !),$ then the value of $\alpha$ is equal to ...... .

If $\mathrm{b}$ is very small as compared to the value of $\mathrm{a}$, so that the cube and other higher powers of $\frac{b}{a}$ can be neglected in the identity $\frac{1}{a-b}+\frac{1}{a-2 b}+\frac{1}{a-3 b} \ldots .+\frac{1}{a-n b}=\alpha n+\beta n^{2}+\gamma n^{3}$, then the value of $\gamma$ is:

If $(1 -x + x^2)^n = a_0 + a_1x + a_2x^2 + ....... + a_{2n}x^{2n}$, then $a_0 + a_2 + a_4 +........+ a_{2n}$ is equal to

The sum of coefficients in the expansion of ${(1 + x + {x^2})^n}$ is

If ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .......... + {C_n}{x^n}$, then $\frac{{{C_1}}}{{{C_0}}} + \frac{{2{C_2}}}{{{C_1}}} + \frac{{3{C_3}}}{{{C_2}}} + .... + \frac{{n{C_n}}}{{{C_{n - 1}}}} = $