The coefficient of $x^{256}$ in the expansion of $(1-x)^{101}\left(x^{2}+x+1\right)^{100}$ is:

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

The value of$^n{C_1}\sum\limits_{r = 0}^1 {^1{C_r}} { + ^n}{C_2}\left( {\sum\limits_{r = 0}^2 {^2{C_r}} } \right){ + ^n}{C_3}\left( {\sum\limits_{r = 0}^3 {^3{C_r}} } \right) + ......{ + ^n}{C_n}\left( {\sum\limits_{r = 0}^n {^n{C_r}} } \right)$ is equal to

If $A$ denotes the sum of all the coefficients in the expansion of $\left(1-3 x+10 x^2\right)^n$ and $B$ denotes the sum of all the coefficients in the expansion of $\left(1+x^2\right)^n$, then :

The sum of the coefficients of three consecutive terms in the binomial expansion of $(1+ x )^{ n +2}$, which are in the ratio $1: 3: 5$, is equal to

$\mathrm{b}=1+\frac{{ }^1 \mathrm{C}_0+{ }^1 \mathrm{C}_1}{1 !}+\frac{{ }^2 \mathrm{C}_0+{ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_2}{2 !}+\frac{{ }^3 \mathrm{C}_0+{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_2+{ }^3 \mathrm{C}_3}{3 !}+\ldots$

Let $\mathrm{a}=1+\frac{{ }^2 \mathrm{C}_2}{3 !}+\frac{{ }^3 \mathrm{C}_2}{4 !}+\frac{{ }^4 \mathrm{C}_2}{5 !}+\ldots$, Then $\frac{2 b}{a^2}$ is equal to.........................