Embibe Experts Solutions for Chapter: Binomial Theorem, Exercise 2: Exercise-2

Let $f\left(n\right)={10}^n+3·4^{n+2}+6,n\in N.$ The greatest value of the integer which divides $f\left(n\right)$ for all $n$ is

Element in set of values of $r$ for which, ${}^{18}C_{r-2}+2·{}^{18}C_{r_{}-1}+{}^{18}C_r\ge {}^{20}C_{13}$ is

The expansion of ${\left(3x+2\right)}^{-\frac{1}{2}}$ is valid in ascending powers of $x,$ if $-x$ lies in the interval.

If ${\left(1+2x+3x^2\right)}^{10}=a_0+a_{1}x+a_2{x}^2+\dots .+a_{20}{x}^{20},$ then

In the expansion of ${\left(x+y+z\right)}^{25}$

If ${\left(1+x+2x^2\right)}^{20}=a_0+a_{1}x+a_2{x}^2+\dots \dots \dots +a_{40}{x}^{40},$ then $a_0+a_2+a_4\dots \dots +a_{38}$ is equal to

$n^n{\left(\frac{n+1}{2}\right)}^{2n}$ is $\left(n\in N\right)$

If recursion polynomials $P_k\left(x\right)$ are defined as $P_1\left(x\right)={\left(x-2\right)}^2, P_2\left(x\right)={\left({\left(x-2\right)}^2-2\right)}^2$ $P_3\left(x\right)={\left({\left({\left(x-2\right)}^2-2\right)}^2-2\right)}^2\dots \dots \dots .$ (In general $P_k\left(x\right)={\left(P_{k-1}\left(x\right)-2\right)}^2,$ then the constant term in $P_k\left(x\right)$ is