The coefficient of $x^{50}$ in the expansion of $(1+x)^{100}+2x(1+x)^{99}+3x^2(1+x)^{98}+\dots+101x^{100}$ is:

Let $(1+x+x^2)^{2014} = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots + a_{4028} x^{4028}$. Let $A = a_0 - a_3 + a_6 - \ldots + a_{4026}$,$B = a_1 - a_4 + a_7 - \ldots - a_{4027}$,and $C = a_2 - a_5 + a_8 - \ldots + a_{4028}$. Then,

Let $a_1, a_2, a_3, \dots, a_{100}$ be positive real numbers and $S_k$ be the sum of products of $a_1, a_2, \dots, a_{100}$ taken $k$ at a time. If $S_{98} S_2 \ge \lambda (a_1 a_2 \dots a_{100})$,then $\lambda$ is

If $C_0, C_1, C_2, \ldots, C_{n}$ are the binomial coefficients in the expansion of $(1+x)^{n}$,then $(C_0+C_1)-(C_2+C_3)+(C_4+C_5)-(C_6+C_7)+\ldots=$

If the coefficient of $x^{30}$ in the expansion of $\left(1+\frac{1}{x}\right)^6\left(1+x^2\right)^7\left(1-x^3\right)^8 ; x \neq 0$ is $\alpha$,then $|\alpha|$ equals

The sum of the series $\sum\limits_{r = 0}^n {(-1)^r \, ^nC_r \left( \frac{1}{2^r} + \frac{3^r}{2^{2r}} + \frac{7^r}{2^{3r}} + \frac{15^r}{2^{4r}} + \dots + m \text{ terms} \right)}$ is