Search results

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| =

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

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

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

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.

$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...

$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...

$Finding largest prime number which divide $\displaystyle \binom{2000}{1000}$ $We can write $\displaystyle \binom{2000}{1000} = \frac{2000!}{1000!\cdot 1000!} = \frac{1001 \cdot 1002 \cdot 1003\cdots \cdots 2000}{1000!}$ i did not understand how to solve further, please explain me

$There are four machines and it is known that exactly two of them are $ $faulty. They are tested, one by one, in a random order till both the faulty $ $machines are identified. Then the probability that only two tests are$ $need is$ $What i have Try:$ $Let $A$ be the event in...

$Total number of positive integer ordered pair of $\binom{a}{b} = 120$ $Using $\binom{a}{b} = 120 = \binom{120}{1} = \binom{120}{119}$. So $(a,b) = (120,1)\;,(120,119)$ $And $\binom{a}{b}$ is $\max$, when $b=\frac{a}{2}$ or $b=\frac{a+1}{2}$ $So must have $b\leq \frac{a}{2}$ or $b \leq...

\displaystyle \bigg\lfloor\frac{2017!}{1!+2!+3!+\cdots \cdots +2016!}\bigg\rfloor $where $\lfloor x \rfloor $ is a floor of $x$ $For upper bound $1!+2!+3!+\cdots \cdots +2016!=2016(1!+2!+3!+\cdots +2016!)$ 2016(1!+2!+3!+\cdots \cdots +2016!)>2016(2016!+2015!)=2017! \displaystyle...

$Evaluation of sum of series $\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\frac{1}{(k-1)\sqrt{k}+k\sqrt{k+1}}$

$If $I = \int^{1}_{0}x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}dx$ and $J = \int^{1}_{0}\frac{x^{\frac{3}{2}}(1-x)^{\frac{7}{2}}}{(3+x)^8}dx$ . Then $\displaystyle \frac{I}{J}$ is

$If $a,b,c$ are three distinct non zero complex numbers such that $ |a| = |b| = |c|$ and the equation $az^2+bz+c=0$ has a roots whose$ $ modulus is $1\;,$ Then relation between $a,b,c$ is$

$Find all parameter $a$ for which the equation $(a-1)4^x-4\cdot 2^x+a+2=0$ $has at least one real solution$