I managed to get $$\sum_{k=0}^n \binom{n+1}{k+1}(-1)^k$$ on the left side, but I don't know how to proceed from here. thanks in advance.
4704
Other than the general inductive method,how could we show that $$\sum_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$$  Apart from induction, I tried with  to check the validity, but I can't think of an easy (manual) alternative.  Please suggest an intuitive/easy method.
110616
The sum of the series $$\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$$   $\bf{My\; Try::}$ We can write it as $\displaystyle \binom{n}{0} = $ Coefficient of $x^0$ in $(1+x)^n$   Similarly $\displaystyle \binom{n-1}{1} = $ Coefficient of $x^1$ in $(1+x)^{n-1}$  Similarly $\displaystyle \binom{n-2}{2} = $ Coefficient of $x^2$ in $(1+x)^{n-2}$  Now, how can I solve it after that, Help Required, Thanks
41502
Calculate  $\sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k}$, I do not know hot get rid of that $k$, for me it is similar like  $\binom{n}{k}=\frac{k}{n} \binom{n-1}{k-1}$, do you have some idea?
106328
Is there a closed form for the following?   $$\sum_{k=0}^{m}  {\binom {m}{k}} a^{k} (b+ck)^N$$      how about a pretty limit for large $b$.   I have tried using the binomial expansion for the $(b+ck)^N$ to convert the above expression to double sum and then tried to simplify the double sum which was not successful. Any idea if I have find a closed form for it?
75307
How can I show that  $$ \sum_{i=1}^{n-1}i=\binom{n}{2}? $$  This is what I have tried, but I do not know if it is correct:  Proof.  Let $n=2$. Then,  $$ \begin{align} \sum_{i=1}^{1}i&amp;=1\text{, and}\\ \binom{2}{2}&amp;=1. \end{align} $$  Hence, it holds. Assume that this is true for $k$. Then, we show that it is also true for $k+1$:  $$ \begin{align} \sum_{i=1}^{k}i=k+\sum_{i=1}^{k-1}i&amp;=k+\binom{k}{2}\\ &amp;=k+\frac{k!}{2(k-2)!}\\ &amp;=k+\frac{k(k-1)}{2}\\ &amp;=\frac{k(k+1)}{2}\\ &amp;=\frac{(k+1)!}{2!(k-1)!}=\binom{k+1}{2}.\square \end{align} $$  Also, proofs by induction are confusing to write. Is there a standard "skeleton" for them I can use?
71474
Prove $$\sum_{k = 0}^{n}(-1)^{n - k} \binom{n}{k} \cdot k^n = n! \qquad\text{and}\qquad \sum_{k = 0}^{n}(-1)^{n - k} \binom{n}{k} \cdot k^m = 0,$$ where $m \in \{0, 1, 2, \cdots, n - 1\}$.  To be honest, I tried to use some combinatorial ideas but was not able to see interconnections. Also, I had an idea to prove it using induction but failed to do that either. Help will be much appreciated! I just got stuck and don't see any continuation of solution for this problem
49569
How to find the sum of this series?  $$\sum_{k=0}^{\infty}\cfrac{{2}^{k}}{\binom{2k+1}{k}}$$  It seems very easy. But I still can not work it out, can anyone help?
22727
Find a combinatorial argument for the following binomial identity:   $$n(n+1)2^{n-2} = \sum_{k=1}^{n}k^2\binom{n}{k} .$$   Algebraic proofs can be found at  , and a related identity at  .
49701
I know that $$\sum_{k=0}^n \binom nk k^2=2^{n-2}(n^2+n),$$ but I cannot find a way how to prove it. I tried induction but it did not work. On wiki they say that I should use differentiation but I do not know how to apply it to binomial coefficient.  Thanks for any response.  (Presumptive) Source: Theoretical Exercise 1.12(b), P18, A First Course in Pr, 8th Ed, by S Ross
89273
$$\binom{n}{0}\binom{n}{1}+\binom{n}{1}\binom{n}{2}+...+\binom{n}{n-1}\binom{n}{n}$$   All I could think of so far is to turn this expression into a sum. But that does not necessarily simplify the expression. Please, I need your help.
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?
59736
How do I see that$$\sum_{k \ge 0} \binom{n}{3k} = (1 + 1)^n + (\omega + 1)^n + (\omega^2 + 1)^n,$$where $\omega = \text{exp}\left({2\over3}\pi i\right)$? What is the underlying intuition behind this equality?
85347
I'm trying to prove the following  $$ \sum_{r=0}^{n}(r+1) \binom{n}{r} = (n+2)2^{n-1} $$  using the identity $$ \sum_{r=0}^{n} \binom{n}{r} = 2^{n} $$  but I'm not able to. This is what I did so far,   $$ \sum_{r=0}^{n}(r+1) \binom{n}{r} = (1) \binom{n}{0} + (2) \binom{n}{1} + ... + (n+1) \binom{n}{n} $$
100910
I was stuck while finding the given summation.  $$\sum_{k=1}^rk^2\binom {n-k}{r-k}$$  Since $n$ and $r$ are both constants, so I have first converted the above summation into this:  $$\sum_{k=1}^rk^2\binom {n-k}{n-r}$$  I have no idea how to proceed next. Any help will be appreciated.
100655
I am to find a combinatorial argument for the following identity:  $$\sum_k \binom {2r} {2k-1}\binom{k-1}{s-1} = 2^{2r-2s+1}\binom{2r-s}{s-1}$$  For the right hand side, I was think that would just be number of ways to choose at least $s-1$ elements out of a $[2r-s]$ set. However, for the left hand side, I don't really know what it is representing.   Any help would be greatly appreciated!
50960
I found interesting task: Calculate $$\sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k}$$ I have calculated some first values and I see that it is $H_n$. I found  tip that it can be solved by induction or by &quot;integral&quot; trick by considering $\sum_{k=1}^n(-1)^k{n\choose k}x^{k-1}$ I don't know what is that trick so I decided to solve it by induction.  Let $S_n = \sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k} $ $$ S_1 = 1 = H_1 \text{ ok.} $$ $$S_{n+1} =  \sum_{k=1}^{n+1} (-1)^{k+1} \binom{n+1}{k}\frac{1}{k}  = \\ -\sum_{k=0}^{n} (-1)^{k+1} \binom{n+1}{k+1}\frac{1}{k+1}$$ but I have problem with use induction assumption. $$-\sum_{k=0}^{n} (-1)^{k+1} \binom{n}{k}\frac{n+1}{(k+1)^2} = \\ -(n+1)\sum_{k=0}^{n} (-1)^{k+1} \binom{n}{k}\frac{1}{(k+1)^2}$$ but know I have $\frac{1}{(k+1)^2} $ instead of something like $\frac{1}{k}$
92914
For $r \geq 1$, $k \geq 0$ both integers, I wish to show that $$\binom{-r}{k}^{*}(-1)^{k} = \binom{r+k-1}{k}$$ (the negative binomial coefficient is the left one). By definition, $$\binom{-r}{k}^{*}(-1)^{k} = (-1)^{k}\dfrac{(-r)(-r-1)\cdots(-r-k+1)}{k!} = (-1)^{k+1}\dfrac{r(r+1)\cdots(r+k-1)}{k!}$$ The right side clearly is $$(-1)^{k+1}\binom{r+k-1}{k}$$ but why does $(-1)^{k+1}$ "disappear"?   See, for example, under .