{n^n}{left( {frac{{n + 1}}{2}} right)^{2n}} is

English

Gujarati

Hindi

7.Binomial Theorem

medium

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

A

Less than ${\left( {\frac{{n + 1}}{2}} \right)^3}$

B

Greater than ${\left( {\frac{{n + 1}}{2}} \right)^3}$

C

Greater than ${(n!)^3}$

D

$(b)$ and $(c)$ both

Solution

(b) $y = {n^n}{\left( {\frac{{n + 1}}{2}} \right)^{2n}}$

Put $n = 2$, $y = {2^2}{\left( {\frac{3}{2}} \right)^4} = 4\,.\,\frac{{81}}{{8 \times 2}} = \frac{{81}}{4}\tilde – 20$

Option $(a) = {\left( {\frac{{n + 1}}{2}} \right)^3} = \frac{{27}}{8} < y$

Option $(b) = {\left( {\frac{{n + 1}}{2}} \right)^3} = \frac{{27}}{8} < y$

Option $(c) = {(2!)^3} = 8 < y$

Standard 11

Mathematics

Share

Similar Questions

For $x\, \in \,R\,,\,x\, \ne \, – 1,$ if ${(1 + x)^{2016}} + x{(1 + x)^{2015}} + {x^2}{(1 + x)^{2014}} + ….{x^{2016}} = \sum\limits_{i = 0}^{2016} {{a_i\,}{\,x^i}} ,$ then $a_{17}$ is equal to

hard

(JEE MAIN-2016)

Find the coefficient of $x^{49}$ in the expansion of $(2x + 1) (2x + 3) (2x + 5)—– (2x + 99)$

normal

The sum of the coefficients in the expansion of ${(1 + x – 3{x^2})^{3148}}$ is

easy

If ${a_k} = \frac{1}{{k(k + 1)}},$ for $k = 1,\,2,\,3,\,4,…..,\,n$, then ${\left( {\sum\limits_{k = 1}^n {{a_k}} } \right)^2} = $

medium

A possible value of $^{\prime}x^{\prime}$, 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:

hard

(JEE MAIN-2021)