Mathematics - Binomial Theorem Question with Solution | TestHub

We note that each option is of the form $(n+1{)}^n>$ or $<n^{n+1}$ Now, $\frac{(n+1{)}^n}{n^{n+1}}=\frac{1}{n}{\left[1+\frac{1}{n}\right]}^n$ But ${\left[1+\frac{1}{n}\right]}^n=1+n\cdot \frac{1}{n}+\frac{n(n-1)}{2!}{\left(\frac{1}{n}\right)}^2+\frac{n(n-1)(n-2)}{3!}{\left(\frac{1}{n}\right)}^3+\dots \dots .$. $=1+1+\frac{1}{2!}\left(1-\frac{1}{n}\right)+\frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+\dots \dots$ Clearly, ${\left(1+\frac{1}{n}\right)}^n>2\dots \dots (1)$ Also ${\left(1+\frac{1}{n}\right)}^n<2+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots \dots$. $\begin{array}{rlrlr}& =2+\frac{\frac{1}{2}}{1-\frac{1}{2}}=3,\text{ provided }n>3& \therefore 2<{\left(1+\frac{1}{n}\right)}^n<3& \Rightarrow \frac{2}{n}<\frac{1}{n}{\left(1+\frac{1}{n}\right)}^n<\frac{3}{n}<1,\text{ for }n>3& \therefore \frac{(n+1{)}^n}{n^{n+1}}<1\Rightarrow (n+1{)}^n<n^{n+1}\text{ for }n>3\end{array}$ Which is satisfied only by the option (b) and (c)