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)
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
$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 \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.