\[\dbinom{2n}{n}=\dfrac{2n}{n}\dfrac{2n-1}{n-1}\dfrac{2n-2}{n-2}\cdots\dfrac{n+2}{2}\dfrac{n+1}{1}\]
Search found 136 matches
- Mon Feb 09, 2015 1:34 am
- Forum: Algebra
- Topic: power of 2 or binomial?
- Replies: 4
- Views: 6646
- Wed Feb 04, 2015 12:52 pm
- Forum: Higher Secondary Level
- Topic: Where did the other root go?
- Replies: 7
- Views: 6125
Re: Where did the other root go?
Yes I do. Then $\dfrac{e^x-e^{-x}}{2}$ changes to $\dfrac{e^{-x}-e^x}{2}$. They're obviously not equal. They'd be if the sign was 'plus'. But we've already dealt with that case.
- Fri Jan 30, 2015 2:00 am
- Forum: Higher Secondary Level
- Topic: Where did the other root go?
- Replies: 7
- Views: 6125
Re: Where did the other root go?
Then how is $-\ln\left(1+\sqrt 2\right)$ another root?
- Mon Jan 26, 2015 9:29 pm
- Forum: Higher Secondary Level
- Topic: Where did the other root go?
- Replies: 7
- Views: 6125
Re: Where did the other root go?
Notice that $e^{2x}-2e^x+1=\left(e^x-1\right)^2=0\Rightarrow e^x=1$ so $x=0$ is the only solution.
Another way to observe is that the left side is the definition of hyperbolic cosine. So $\cosh x=1$ gives $x=2\pi i n$ for all $n\in\mathbb{Z}$. Only real solution is $x=0$ for $n=0$.
Another way to observe is that the left side is the definition of hyperbolic cosine. So $\cosh x=1$ gives $x=2\pi i n$ for all $n\in\mathbb{Z}$. Only real solution is $x=0$ for $n=0$.
- Mon Jan 26, 2015 1:12 am
- Forum: Number Theory
- Topic: China TST 2008
- Replies: 2
- Views: 3088
Re: China TST 2008
For any prime $p$ dividing $n$ we have $\displaystyle \sum_{k=2}^n k^{\phi(n)}\equiv n-\dfrac{n}{p}-1~(\bmod~ p)$ because there are $\dfrac{n}{p}$ multiples of $p$ in $[2,n]$, thus $n-\dfrac{n}{p}-1$ integers not divisible by $p$, which are congruent to $1~(\bmod~ p)$ by Euler's theorem. Hence by hy...
- Wed Jan 21, 2015 10:42 pm
- Forum: Combinatorics
- Topic: Partial Generalization of ISL '94 C5
- Replies: 5
- Views: 4692
Re: Partial Generalization of ISL '94 C5
Sorry, typo, fixed.
- Wed Jan 21, 2015 3:03 pm
- Forum: Combinatorics
- Topic: Partial Generalization of ISL '94 C5
- Replies: 5
- Views: 4692
Re: Partial Generalization of ISL '94 C5
That's why I said *partial*.
- Tue Jan 20, 2015 2:02 pm
- Forum: Combinatorics
- Topic: n+1 rows and columns
- Replies: 10
- Views: 8036
Re: n+1 rows and columns
If there existed an injective map, subset of the original map, that would mean there are $n$ cells none of which lies on same row/column, are all white. For example, consider the following. http://s23.postimg.org/8ji7lt1y3/Map.png Here we have an injective map in the left, showing the cells $(1, 1),...
- Tue Jan 20, 2015 1:34 pm
- Forum: Combinatorics
- Topic: Partial Generalization of ISL '94 C5
- Replies: 5
- Views: 4692
Partial Generalization of ISL '94 C5
Suppose $2n$ girls are seated around a circle $(n\in\mathbb N).$ One of the girls is given $2n$ coins initially. In a move, each girl with at least $2$ coins passes one coin to each of her two neighbors. The game terminates if, in a move, no girl is able to pass any coins. Prove that the game can't ...
- Tue Jan 20, 2015 2:31 am
- Forum: Combinatorics
- Topic: n+1 rows and columns
- Replies: 10
- Views: 8036
Re: n+1 rows and columns
I considered the problem to be much trivial. :? Call the cell on row $i$ and column $j$ by $(i,j)$. Consider a visual map (or relation) $F$ of two sets $\mathcal A,\mathcal B$ each containing the integers from $1$ to $n$. We connect two integers $a\in\mathcal A$ and $b\in \mathcal B$ by a segment if...