Embibe Experts Solutions for Chapter: Binomial Theorem, Exercise 3: Level 3

The lowest integer which is greater than ${\left(1+\frac{1}{{10}^{100}}\right)}^{{10}^{100}}$ is

A possible value of $x,$ for which the ninth term in the expansion of ${\left\{3^{{\log }_3\sqrt{{25}^{x-1}+7}}+3^{\left(-\frac{1}{8}\right){\log }_3\left(5^{x-1}+1\right)}\right\}}^{10}$ in the increasing powers of $3^{\left(-\frac{1}{8}\right){\log }_3\left(5^{x-1}+1\right)}$ is equal to $180,$ is :

Let $a_k$ be the coefficient of $x^k$ in the expansion of ${\left(1+2x\right)}^{100}$, where $0\le k\le 100.$ Find the number of integers $r:0\le r\le 99$ such that $a_r<a_{r+1}$.

If $n$ be a positive integer such that $n\ge 3,$ then the value of the sum upto $n$ terms of
$1·n-\frac{(n-1)}{1!}\left(n-1\right)+\frac{(n-1)(n-2)}{2!}\left(n-2\right)-\frac{(n-1)(n-2)(n-3)}{3!}\left(n-3\right)+\dots \dots$ is

The value of the sum $\sum _{j=0}^8\left[\frac{1}{\left(j+1\right)\left(j+2\right)}\left({}^{8}C_j\right)\right]$ is

For natural numbers $m \& n$, if ${\left(1-y\right)}^m{\left(1+y\right)}^n=1+a_{1}y+a_2{y}^2+\dots$ and $a_1=a_2=10,$ then $\left(m,n\right)$ is

In the binomial expansion of $(1+x{)}^{2k},$ if its middle term is the only numerically greatest term, then $x$ lies in the interval

For $x>0,$ if $p^{\text{th }}$term is the first negative term in the expansion of ${\left(1+\frac{3x}{5}\right)}^{22/3}$and in the expansion of ${\left(1-\frac{3x}{5}\right)}^{22/3}$ from $r^{th}$term onwards all the terms are positive, then the number of terms in the expansion of ${\left(px+\frac{r}{x}\right)}^{pr}$ is