I have:   $$\sum\limits_{k = 1}^{10}k\binom{10}{k}\binom{20}{10-k} = $$  and I know that it doesn't matter if $k = 0$ so it also equals:  $$= \sum\limits_{k = 0}^{10}k\binom{10}{k}\binom{20}{10-k} = $$  and now it really reminds Vandermonde's identity so it's tempting to write:  $$= \binom{30}{10} \cdot \sum\limits_{k = 1}^{10}k = \binom{30}{10} \cdot 10!$$  but it seems wrong because on smaller numbers the equations doesn't hold... What's the right way to continue in order to get an equivalent expression without sigma?
122267
I still haven't quite realized how to solve binomial coefficient problems like this, can someone show me an elaborated way of solving this?  I need to write this expression in a more simplified way: $\sum_{k=0}^{10} k \pmatrix{10 \\ k}\pmatrix{20 \\ 10-k}$  I'm aware of the binomial identity $\sum_{k=i}^{n-j}\binom{k}{i}\binom{n-k}{j}=\binom{n+1}{j+1}$  Just can't seem to see how to use it.  Many thanks.
114867
How to get a closed form for $$\sum_{k=0}^{n-p}\binom{n}{k}\binom{n}{p+k}\;?$$  I tried to write binominal in term of gamma function but I got no result. What is your suggestion to solve the problem?
4231
I am being asked to prove that $$\sum\limits_{k=0}^{n}\cos(kx)=\frac{1}{2}+\frac{\sin(\frac{2n+1}{2}x)}{2\sin(x/2)}$$ I have some progress made, but I am stuck and could use some help. What I did: It holds that $$\sum\limits_{k=0}^{n}\cos(kx)=\sum\limits_{k=0}^{n}Re(\cos(kx))=\sum\limits_{k=0}^{n}Re(\cos(x)^{k})=Re(\sum\limits_{k=0}^{n}\cos(x)^{k})=Re\left(\cos(0)\cdot\frac{\cos(x)^{n}-1}{\cos(x)-1}\right)=Re\left(\frac{\cos(x)^{n}-1}{\cos(x)-1}\right) $$ For any $z_{1},z_{2}\in\mathbb{C}$ we have it that if $z_{1}=a+bi,z_{2}=c+di$ then $$\frac{z_{1}}{z_{2}}=\frac{z_{1}\overline{z2}}{|z_{2}|^{2}}=\frac{(a+bi)(c-di)}{|z_{2}|^{2}}=\frac{ac-bd+i(bc-ad)}{|z_{2}|^{2}}$$ hence $$Re\left(\frac{z_{1}}{z_{2}}\right)=\frac{Re(z_{1})Re(z_{2})-Im(z_{1})Im(z_{2})}{|z_{2}|^{2}}$$ Thus, $$Re\left(\frac{\cos(x)^{n}-1}{\cos(x)-1}\right)=\frac{(\cos(nx)-1)(\cos(x)-1)-\sin(nx)\sin(x)}{(\cos(x)-1)^{2}+\sin^{2}(x)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{\cos^{2}(x)-2\cos(x)+1+\sin^{2}(x)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2\cos(x)+2}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2(\cos(x)-1)}= \frac{=\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2(-2\cdot\sin^{2}(x/2))}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{4\sin^{2}(x/2)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{4\sin^{2}(x/2)}=\frac{\cos(x(n+1))-\cos(nx)-\cos(x)+1}{4\sin^{2}(x/2)} $$ This is the part where I am stuck, I would appriciate any help or hint on how to continue. Edit: Given the corrections by André I get: $$(\cos(nx+x)-1)(\cos(x)-1)+\sin(nx+x)\sin(x)=\cos(nx+x)\cos(x)-\cos(nx)-\cos(x)+1+\sin(nx+x)\sin(x)$$ so $$\cos(nx+x)\cos(x)+\sin(nx+x)\sin(x)=\cos(xn+x-x)-\cos(nx)=0$$ Edit 2: I found anoter mistake in the above, I will try to correct Edit 3: When multiplying correctly the above it works out :-)
6574
How to simplify $\sum_{i=1}^{k}\binom{n + i - 1}{i}$? I tried reducing the sum to $\binom{n}{1}, \binom{n}{2}, \binom{n}{3}$ and so on but couldn't get a pattern.
118721
I'm undergraduate student of mathematics.  I need to prove: $$\sum_{k=0}^{n} \binom{2n}{2k}= 2^{2n-1}$$ Can you please help me
16808
While thinking about  I found out that this equality might be useful there: $$n\cdot 1 + (n-1)\cdot 2 + \dots + 2\cdot (n-1) + 1\cdot n = \frac{n(n+1)(n+2)}6$$ To rewrite it in a more compact way: $$\sum_{k=1}^n k(n+1-k)=\frac{n(n+1)(n+2)}6.$$  This equality is relatively easy to prove: $$\sum_{k=1}^n k(n+1-k)= (n+1)\sum_{k=1}^n k - \sum_{k=1}^n k^2 = (n+1) \frac{n(n+1)}2 - \frac{n(n+1)(2n+1)}6  = n(n+1) \left(\frac{n+1}2-\frac{2n+1}6\right) = n(n+1)\frac{3(n+1)-(2n+1)}6 =  \frac{n(n+1)(n+2)}6.$$ (We only used the known formulas for the   and the .)   Are there some other nice proofs of this equality? (Induction, combinatorial arguments, visual proofs, ...)     EDIT: Now I found another question which asks about the same identity:  (I have tried to search before posting. But the answers posted here so far gave me some new ideas for good keywords to search which lead me to finding that question.) The questions are, in my opinion, not exact duplicates since the other question asks specifically about combinatorial proofs and my question does not have that restriction. But I agree that this is a very minor distinction. In any case, if you think that one of them should be closed as a duplicate, then you can vote to close. I will refrain from voting to close/reopen on this question. (If one of the two questions is voted to be a duplicate of the other one, they probably cannot be merged, since the summation variables are off by one.)
109085
How can you show that $$ \binom{2n}{n}^2 = \sum_{m=0}^{n} \binom{2n}{2m} \binom{2m}m \binom{2n-2m}{n-m} $$?  I was fooling around with random walks, and apparently both expressions are supposed to be the number of unweighted random walks on a 2D square lattice of $2n$ steps that ends at the origin.  Whole thing reminds me of Vandermonde's identity but I can't it. Any hints?
41291
How do I go about simplifying this:  $$\sum_{k=1}^nk \cdot k!$$  Wolfram alpha tells me it's the same as $(n+1)!-1$ but I don't see how.
83623
I'm trying to prove the following equation above. So far I have: \begin{align} 2^{2n} &amp;= (1+1)^{2n}\\ &amp;= \sum_{k=0}^{2n}\binom{2n}{k}1^k1^{n-k} = \sum_{k=0}^{2n}\binom{2n}{k} &amp; \text{(By the Binomial Theorem)} \end{align}  I know I have to use the following identity somehow: $$\binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k}$$  How do I split my summation to get what I'm looking for? Thanks!  EDIT: HERE IS MY SOLUTION \begin{align*} 2^{2n} &amp;= (1+1)^{2n}\\ &amp;= \sum_{k=0}^{2n}\binom{2n}{k}1^k1^{2n-k} &amp; \text{(By the Binomial Theorem)}\\ &amp;= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=n+1}^{2n}\binom{2n}{k}\\ &amp;= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=n+1}^{2n}\binom{2n}{2n-k} &amp; \text{(Binomial Symmetry)}\\ &amp;= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=0}^{n-1}\binom{2n}{k}\\ &amp;= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=1}^{n}\binom{2n}{k-1}\\ &amp;= \binom{2n}{0} + \sum_{k=1}^{n}\binom{2n}{k} + \sum_{k=1}^{n}\binom{2n}{k-1}\\ &amp;= \binom{2n}{0} + \sum_{k=1}^{n}\binom{2n+1}{k} &amp; \text{(By Identity listed above)}\\ &amp;= \sum_{k=0}^{n}\binom{2n+1}{k} \end{align*}
27537
closed form for  $$\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+...+\binom{n}{n}$$  I tried to solve it by :   $$\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+...+\binom{n}{n}=\sum_{k=0}^{n/3}\binom{n}{3k}$$  $$\sum_{k=0}^{n/3}\binom{n}{3k}=\sum_{k=0}^{n/3}\frac{1}{2\pi i}\int_{\left |z  \right |=1}\frac{(1+z)^n}{z^{3k+1}}dz$$  then using geometric series but i got no result .  what is your suggest to solve ?
104266
Please give me feedback on my answer to this question. Question: For all $ n\geq1:\binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n} $   is equal to $ \binom{2n}{1}+\binom{2n}{3}+\binom{2n}{5}+\cdots+\binom{2n}{2k+1}+\cdots+\binom{2n}{2n-1} $. Answer: False, since;  Let $a=1,  b=-1$  and $n=1$  ,$k=0$   , using Binomial expression.  $$(a+b)^n=  \sum   \binom{n}{k}  a^{k}b^{n-k}$$  Then $(1+-1)^{2(1)}=\binom{2}{0}1^{0}(-1)^{2-0}+\binom{2}{1}1^{1}(-1)^{2-1}+\binom{2}{2}1^{2}(-1)^{2-2}=\binom{2}{0}-\binom{2}{1}+\binom{2}{2}$  Thus :For all $n\geq1:  \binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n}   \ne \binom{2n}{1}+\binom{2n}{3}+\binom{2n}{5}+\cdots+\binom{2n}{2k+1}+\cdots+\binom{2n}{2n-1}$  .
10614
Show that $$\sum_{k=0}^n(-1)^k\binom nk=0$$ So for odd $n$ we have an even number of terms. So $\binom nk=\binom n{n-k}$ which have opposite signs. Thus the sum is 0.  For even $n$ we have that $$\sum_{k=0}^n(-1)^k\binom nk= \binom n0+\sum_{k=1}^{n-1}(-1)^k\binom nk+\binom nn$$ Now $$\sum_{k=1}^{n-1}(-1)^k\binom nk= \sum_{k=1}^{n-1}(-1)^k\left[\binom{n-1}k+\binom{n-1}{k-1}\right]$$ What would that sum be in the square brackets?
110367
I'm trying to simplify the following formula, but I keep failing to do.   $$ \sum_{k=0}^{\infty} \binom{k + r - 1}{k} x^k $$  I tried:  From the Hockey-stick identity,  $$ \sum_{k=0}^{n} \binom{k + r - 1}{k}  = \binom {n+r}{n}$$  But I have no clue to apply the fact to simplify the above formula.  Is there anyone to give a hint for it?
98822
Find the value of $\binom{2000}{2} + \binom{2000}{5} + \binom{2000}{8} + \cdots \binom{2000}{2000}$     I've seen many complex proofs. I am looking for an elementary proof. I know the fact that $\binom{2000}{0} + \binom{2000}{1} + \binom{2000}{2} + \cdots \binom{2000}{2000} = 2^{2000}$. This may help here.
26130
Why is $\sum_{k=0}^n (_k^n) 2^k$ simplified as $3^n$ and not just $2^n$ ? The answer uses the binomial theorem as a solution: let $a=1$ and let $b=2$. So $1+2 = 3$. But since $1+2 = 3$ why do we not use $1+1 = 2$?
52285
I am trying to get a combinatorial proof for this summation identity, for $0\leq k\leq \frac{n}{2}$ $$\sum_{m=k}^{n-k}\binom{m}{k}\binom{n-m}{k}=\binom{n+1}{2k+1}$$  The way I interpreted is that, for the right hand side $\binom{n+1}{2k+1}$ is the number of ways to choose $2k+1$ from total of $n+1$, then for left hand side, condition on dividing $n+1$ into two groups of size $m+1$ and $(n+1)-(m+1)=n-m$, then if choose $k$ the group of $n-m$ then it gives $\binom{n-m}{k}$, so there are $2k+1-k=k+1$ to choose from the group of size $m+1$, this is $\binom{m+1}{k+1}$, therefore I got the left hand side as $\sum_{m=k}^{n-k}\binom{m+1}{k+1}\binom{n-m}{k}\neq\sum_{m=k}^{n-k}\binom{m}{k}\binom{n-m}{k}$. what did I do wrong here? please help.
7233
How can I prove the identity:  $$ \sum_{k=0}^n\binom{2n}{2k}^2-\sum_{k=0}^{n-1}\binom{2n}{2k+1}^2=(-1)^n\binom{2n}{n}? $$  Maybe, can we expand $$ f(x)=(1+x)^{2n}? $$  Thank you.