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Evaluation of the sum \displaystyle \frac{\sum^{r}_{k=0}\binom{n}{2k}\cdot \binom{n-2k}{r-k}}{\sum^{n}_{k=r}\binom{n}{k}\cdot \binom{2k}{2r}(0.75)^{n-k}\cdot (0.5)^{2k-2r}}\ \ ,\ \ (n\geq 2r)

A have a doubt : If \displaystyle A=[a_{ij}]_{0\times 0} is a matrix . Then \displaystyle |A| =

Thanks Tywebb. I have tried like this way I am assuming \displaystyle n=2k, k\geq 3, k\in\mathbb{Z} So we have \displaystyle \frac{n!}{\bigg(\frac{n}{2}\bigg)^n}=2\prod^{\frac{n}{2}-1}_{r=1}\frac{\bigg(\frac{n}{2}-r\bigg)\bigg(\frac{n}{2}+r\bigg)}{\frac{n}{2}\cdot...

Thanks cossine. Using your hint , We use Stirling approximation \displaystyle n!\approx \bigg(\frac{n}{e}\bigg)^{n}\sqrt{2\pi n}\approx \bigg(\frac{n}{e}\bigg)^n. So we have \displaystyle \frac{n}{\bigg(n!\bigg)^{\frac{1}{n}}}\approx e>2\Longrightarrow n!<\bigg(\frac{n}{2}\bigg)^n

Proving the result \displaystyle \frac{n^n}{n!}>2^n, n>6

Thanks Tywebb.

Thanks Tywebb. Can we solve it without using Stolz and Stirling approximation like definite integration If yes then please explain me.

Evaluation of \displaystyle \lim_{n\rightarrow \infty}\bigg(((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}}\bigg)

Thanks Tywebb for different Methods.

Evaluation of \displaystyle \lim_{n\rightarrow \infty}\bigg[\binom{n}{0}\cdot \binom{n}{1}\cdot \binom{n}{2}\cdots \cdots \binom{n}{n}\bigg]^{\frac{1}{n(n+1)}}

(a)\; $Evaluation of $\int\frac{\sqrt{9x^2+4x+6}}{8x^2+4x+6}dx$ (b)\; $Evaluation of $\int\frac{x^2(x\sec x+\tan x)}{(x\tan x-1)^2}dx.$

$Consider a sequence $b_{n}$ given as $b_{1}=\frac{1}{3}$ $ and $b_{n+1}=b^2_{n}+b_{n}$. Then value of $\lfloor \sum^{2008}_{k=2}\frac{1}{b_{k}}\rfloor.

$Evaluation of $\int^{\frac{\pi}{2}}_{0}\cos x\cdot \ln(\cos x)dx

Re: HSC 2018 MX2 Integration Marathon $Let $I = \int^{1}_{0}\frac{\ln(1+x)}{1+x^2}dx,$ $Put $x=\frac{1-t}{1+t}=\frac{2}{t+1}-1\;,$ Then $dx=-\frac{2}{(1+t)^2}dt $And changing limits$ $So $I = \int^{1}_{0}\frac{\ln(2)-\ln(1+t)}{1+t^2}dt=\int^{1}_{0}\frac{\ln(2)}{1+t^2}dt-I$ $So we have $2I...

Evaluation of \displaystyle \int\frac{x^5+3}{(x-5)^3\cdot (x-1)}dx Although i have solved it using partial fraction. But it is very lengthy so please explain me any bettter method. Thanks.

To paradoxica would you like to explain me in detail. Thanks

$Finding value of $ $\displaystyle \lim_{n\rightarrow \infty} n^{-2}\cdot \sum^{n}_{k=0}\ln\bigg(\binom{n}{k}\bigg)$

$ Finding value of $ (0.5+\cos \alpha)\times(0.5+\cos 3\alpha)\times(0.5+\cos 9\alpha)\times(0.5+\cos 27\alpha)$ $ Where $\alpha =9^\circ}$ I have tried like this way $ let $(x+\cos \alpha)(x+\cos 3\alpha)(x+\cos 9\alpha)(x+\cos 27\alpha)$ $which equals $x^4+(\sum \cos \alpha)x^3+(\cos...

Thanks camelrider.

$Let $\displaystyle a_{n} = n+\frac{1}{n}$ for $n=1,2,3,4,....20$ and $p=\frac{1}{20}\sum^{20}_{n=1}a_{n}$ $And $q = \frac{1}{20}\sum^{20}_{n=1}\frac{1}{a_{n}}$. Then proving $q\in \left(0,\frac{21-p}{21}\right)$. Trial p = \frac{1}{20}\left((1+2+3+\cdot +20)+1+\frac{1}{2}+\frac{1}{3}+\cdots...