I know that: $${n \choose 0} + {n \choose 1} + ... + {n \choose n} = 2^n.$$  Does $$0 {n \choose 0} + 1 {n \choose 1} + ... + n {n \choose n} = ??$$  have some convenient simplification?
9697
I am trying to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$ but am not able to think something except induction,which is of-course not necessary (I think) here, so I am inquisitive to prove this in a more general way.  The left side of an identity occurs while solving another problem (concerning binomial theorem) so I am more interested in deriving the right side from the left side, else I have to remember it now onward.  EDIT: I am more interested in an algebraic proof rather than combinatorial argument or something involving calculus (however I liked  and  solution), hence I am removing the combinatorics tag.
118689
Could you help me to find an asymptotic for this sum? $$ \sum_{k=0}^{n - 1} (-1)^k {n \choose k} {3n - k - 1 \choose 2n - k}  = {n \choose 0} {3n - 1 \choose 2n} - {n \choose 1} {3n - 2 \choose 2n - 1}  + ... + (-1)^{n-1} {n \choose n-1} {2n \choose n + 1}  $$  I have tried to write binomials through factorials and work with it, but it seems like not right way to evaluate an asymptotic.  Thank you for your help:)
45648
Since everyone freaked out, I made the variables are the same. $$ \sum_{x=1}^{n} 2^{x-1} $$  I've been trying to find this for a while. I tried the usually geometric equation  but I couldn't get it right (if you need me to post my work I will). Here's the outputs I need:  1, 3, 7, 15, 31, 63  If my math is correct.
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 $1+1 = 2$?
43106
I do as follows  $(1+x)^n= \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+...+\binom{n}{n}x^n$   differentiating in both sides  $n(1+x)^{n-1}= \binom{n}{1}+2\binom{n}{2}x+3\binom{n}{3}x^2...+n\binom{n}{n}x^{n-1}$  then, for $x=1$  $n \cdot2^{n-1} = \sum_{k=1}^{n}k\binom{n}{k}$  From here, I don't know how to proceed, I have tried to multiply both sides for $(1+x)$ to make $n \cdot2^{n}$ but then the summatory changes, I don't know how to prove the equality.
31453
I'm well aware of the combinatorial variant of the proof, i.e. noting that each formula is a different representation for the number of subsets of a set of $n$ elements. I'm curious if there's a series of algebraic manipulations that can lead from $\sum\limits_{i=0}^n \binom{n}{i}$ to $2^n$.
103880
Just checking. $2^n$ ($n \to \infty$) tends to $\infty$. + $3^n$ also ($n \to \infty$) tends to $\infty$  so the sum gets me $\infty$.  Now $(\infty)^{1/\infty}$ : $(\infty)^0 = 1$  I see no other way.  Theorem: $n^{\frac{1}{n}} = 1$ as $n \to \infty$.  Analogous:  $(2^n - 3^n)^\frac{1}{n}$ = $(\infty  - \infty)^0$ = 1 ???  Is it ok to suppose infinity on any integer $k^n$ as $n$ goes to infinity ?
70000
I have to prove that:     $$\sum_{k=0}^{n} {{m+k} \choose{m}} = { m+n+1 \choose m+1 }$$   I tried to open up the right side with Pascal's definition that: $$ { n \choose k} = {n-1 \choose {k}} + {n-1 \choose {k-1}}$$  Here is what I came up with, and I am sure it is wrong because it does not equal the left side:  $$ {m+n+1 \choose m+1} = {m+n \choose m+1} + {m+n \choose m} = ... ={m+n \choose m+1} + {m+n-1 \choose m} + {m+n-2 \choose m-1} + ... + {m \choose m+1} = \sum_{k=0}^{n} {m+k \choose m+k-n+1 }  $$  Which, again, probably is wrong because it is not equal $\sum_{k=0}^{n} { m+k \choose m}$. Any help is appreciated
100263
I'm trying to show that      $$\sum_{k=0}^n (-1)^k\frac{{ {n}\choose{k}}}{{ {x+k}\choose{k}}} = \frac{x}{x+n}$$   I've tried expanding this using definition of binomial coefficient, then binomial expansion, now I'm using induction to simplify $\sum_{k=0}^m (-1)^k\frac{{ {n}\choose{k}}}{{ {x+k}\choose{k}}}$, but it looks very inelegant, is there a better way?
90426
I need to show that $$\sum\limits_{k=i}^{n}\binom{n}{k}\binom{k}{i} = \binom{n}{i}2^{n-i}$$  I know that $\displaystyle \binom{n}{k}\binom{k}{i}$ is counting the number of ways to pick $k$ elements from $n$ and then $i$ elements from those $k$, so $\displaystyle \sum\limits_{k=i}^{n}\binom{n}{k}\binom{k}{i}$ is counting the number of ways to do that $\forall k$.  I know that $\displaystyle \binom{n}{i} = \binom{n}{n-i}$ and that $2^{n-i}$ is the number of elements in the power set for $[n]\setminus[i]$.  How do I show that both sides are equivalent? I'm stumped on this one. Could anyone give me a pointer in the right direction?
90968
Must show:     $\sum_{i=0}^n i{n \choose i} = n2^{n-1}$ using a combinatorial argument. I have tried picking the problem apart by determining that the summation is equal to 1/2 of the size of all subsets of a set of size n times n.  But I am having trouble relating that to what each term of the summation represents.
51603
Prove the identity:  $$n(n-1)2^{n-2}=\sum_{k=1}^n {k(k-1) {n \choose k}}$$  I tried using the binomial coefficients identity $2^n = \sum_{k=1}^n {n \choose k}$  but got stuck along the way.
52105
For any positive integer $n$ (using integer division only), let $P(n)$ denote the number of ways in which the integer $(n^2+n)/4$ can be expressed as a sum of exactly $n /2$ distinct elements of the set $\{1,2,3,\dots, n\}$.    What is $P(n)$ in terms of n?  Specifically, how exponential is it? Is this less than $2^{n/2}$?
124091
No. of ways to choose $n$ things from $n$ alike "a" things $n$ alike "b" things and $n$ different things.  Answer is $(n+2)2^{n-1}$. But how to prove it..?
87923
Prove that $$\sum_{k=0}^n {n \choose k}^{\!2} = {2n \choose n}.$$  The exercise provides the following hint: $\,\,\displaystyle{n \choose k}={n\choose n-k}$.  Any help?
6479
For $n\in\mathbb{N}^*$ we note $s(n)$ the sum of the digits of $2^n$. For example $s(4)=7$ because $2^4=16$. Is it possible to find an integer $n$ such that $s(n)=s(n+1)$?
88921
Can this identity be derived from the binomial theorem?  $$n(n+1)2^{n-2} = \sum_{i=1}^{n}  i^2 \binom{n}{i}$$    I tried starting from $2^n = \displaystyle\sum_{i=0}^{n} \binom{n}{i}$ and dividing it by $4$ in order to get $2^{n-2}$.