For $x\, \in \,R\,,\,x\, \ne \, – 1,$ if ${(1 + x)^{2016}} + x{(1 + x)^{2015}} + {x^2}{(1 + x)^{2014}} + ….{x^{2016}} = \sum\limits_{i = 0}^{2016} {{a_i\,}{\,x^i}} ,$ then $a_{17}$ is equal to

Let $[ x ]$ denote greatest integer less than or equal to $x .$ If for $n \in N ,\left(1-x+x^{3}\right)^{n}=\sum_{j=0}^{3 n} a_{j} x^{j}$, then $\sum_{j=0}^{\left[\frac{3 n}{2}\right]} a_{2 j}+4 \sum_{j=0}^{\left[\frac{3 n-1}{2}\right]} a_{2 j+1}$ is equal to

The coefficient of $x^{91}$ in the series $^{100}{C_1}\,{2^8}.\,{\left( {1\, – \,x} \right)^{99}}\, + {\,^{100}}{C_2}\,{2^7}.\,{\left( {1\, – \,x} \right)^{98}}\, + {\,^{100}}{C_3}\,{2^6}.\,{\left( {1\, – \,x} \right)^{97}}\, + \,….\, + {\,^{100}}{C_9}\,{\left( {1\, – \,x} \right)^{91}}$ is equal to –

If $C_{x} \equiv^{25} C_{x}$ and $\mathrm{C}_{0}+5 \cdot \mathrm{C}_{1}+9 \cdot \mathrm{C}_{2}+\ldots .+(101) \cdot \mathrm{C}_{25}=2^{25} \cdot \mathrm{k}$ then $\mathrm{k}$ is equal to

Let $X =\left({ }^{10} C _1\right)^2+2\left({ }^{10} C _2\right)^2+3\left({ }^{10} C _3\right)^2+\ldots \ldots . .+10\left({ }^{10} C _{10}\right)^2$ where ${ }^{10} C _{ r }, r \in\{1,2, \ldots ., 10\}$ denote binomial coefficients. Then, the value of $\frac{1}{1430} X$ is. . . . . . .