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

The sum of the series $\left(1^2+1\right)·1!+\left(2^2+1\right)·2!+\left(3^2+1\right)·3!+...+\left(n^2+1\right)·n!$ is

The binomial expansion of ${\left(x^k+\frac{1}{x^{2k}}\right)}^{3n}, n\in N$ contains a term independent of $x$

Let $n\in N$. If ${\left(1+x\right)}^n=a_o+a_{1}x+a_2{x}^2+....+a_n{x}^n$ and $a_{n-3}, a_{n-2}, a_{n-1}$ are in $A.P$, then

If the coefficients of three consecutive terms in the expansion of ${\left(1+x\right)}^n$ are in ratio of $1:7:42,$ then $n$ is divisible by

If ${\left(1+x+x^2+x^3\right)}^{100}=a_0+a_{1}x+a_2{x}^2+...+a_{300}{x}^{300}$, then

If ${\left(9+\sqrt{80}\right)}^n=\mathrm{l}+f$ where $\mathrm{l},n$ are integers and $0<f<1$, then

In the expansion of ${\left(x^{\frac{2}{3}}-\frac{1}{\sqrt{x}}\right)}^{30}$, a term containing the power $x^{13}$

The co-efficient of the middle term in the expansion of ${\left(1+x\right)}^{2n}$ is