$\sum_{\substack{i, j=0 \\ i \neq j}}^{n} {}^{n}C_{i} {}^{n}C_{j}$ is equal to

The value of $\sum\limits_{n = 0}^4 {{{\left( {1009 - 2n} \right)}^4} \binom{4}{n} {\left( { - 1} \right)^n}}$ is

Let $S_n = 1 + q + q^2 + ..... + q^n$ and $T_n = 1 + \left( \frac{q + 1}{2} \right) + \left( \frac{q + 1}{2} \right)^2 + ...... + \left( \frac{q + 1}{2} \right)^n$ where $q$ is a real number and $q \ne 1$. If $^{101}C_1 + ^{101}C_2 \cdot S_1 + ...... + ^{101}C_{101} \cdot S_{100} = \alpha \cdot T_{100}$,then $\alpha$ is equal to

If the coefficient of $x^r$ in the expansion of $(1+x+x^2+x^3)^{100}$ is $a_r$,and $S = \sum_{r=0}^{300} a_r$,then $\sum_{r=0}^{300} r \cdot a_r =$

If $(1 + x) (1 + x + x^2) (1 + x + x^2 + x^3) \dots (1 + x + x^2 + \dots + x^n) \equiv a_0 + a_1x + a_2x^2 + a_3x^3 + \dots + a_mx^m$,then $\sum_{r=0}^m a_r$ has the value equal to:

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,