Prove by induction that   $$\sum_{i=0}^n\binom ni=2^n\text{ where }\binom ni=\frac{n!}{i!(n-i)!}$$   Please explain me step by step. Thank you.
25221
Prove by induction that for all $n \ge 0$:  $${n \choose 0} + {n \choose 1} + ... + {n \choose n} = 2^n.$$  In the inductive step, use Pascal’s identity, which is:  $${n+1 \choose k} = {n \choose k-1} + {n \choose k}.$$  I can only prove it using the binomial theorem, not induction.
66163
I ran across this problem in some old notes, and I frustratingly can't figure out how to do it   Let $a_i$ and $b_i$ be sequences of natural numbers, use induction to show  $\sum_{i=1}^n (a_ib_i)^{1/2} \le (\sum_{i=1}^n a_i)^{1/2}(\sum_{i=1}^n b_i)^{1/2} $  Obviously this is trivial to show for n=1. I can't make much progress on n+1. I've tried various tactics, squaring both sides etc.   Any hint or help would be appreciated. Thanks.
39049
Given the identity  $$\sum_{i=0}^{n} \binom{k-1+i}{k-1} = \binom{n+k}{k}$$  Need to give a combinatorial proof   a) in terms of subsets  b) by interpreting the parts in terms of compositions of integers  I should not use induction or any other ways...  Please help.
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,   $$ \sum_{r=0}^{n}(r+1) \binom{n}{r} = (1) \binom{n}{0} + (2) \binom{n}{1} + ... + (n+1) \binom{n}{n} $$
88351
I would like to prove the following inequality, for $n=0,1,2,...$,   $$ \binom{2n}{n}\leq 4^n.$$ I already proved it by induction, and I'm looking for another proof.
123943
$$n!=\sum_{k=0}^n \binom{n}{k}(n-k+1)^n(-1)^k$$  Could anyone give the proof of the above equation? Thanks in advance!
104791
I would like your help with proving that for every  $0 \leq k \leq n$,      $$\binom{n}{k}^{-1}=(n+1)\int_{0}^{1}x^{k}(1-x)^{n-k}dx  . $$   I tried to integration by parts and to get a pattern or to use the binomial formula somehow, but it didn't go well.  Thanks a lot!
33723
Question:  $$\sum_{k=1}^{n}k\binom{2n}{n+k}=\frac n2\binom{2n}{n}$$ is true for every $n\in \mathbb N$? If this is true, then how can we prove this?   When I was playing with numbers, I conjectured this. Though I found that this holds for many $n$s, I haven't found any proof. Can anyone help?
65822
Please feel free to close this is necessary as I didn't see exactly this question (some variations that I tried but didn't seem to apply. Prove: $$\sum_{k=0}^{n}{\binom{n}{k}^2}=\binom{2n}{n}$$ I figured trying to muscle through it by expanding and then multiplying each term by variants of 1 such as $\frac{n^2}{n^2}$ and $\frac{n^2(n-1)^2}{n^2(n-1)^2}$ to get common denominators, but that was NOT the way it seemed as it got ugly pretty quickly.  Then I tried thinking up formulas of the binomial theorem that I could differentiate like other sums of binomial coefficient problems and i couldn't think of one.  Any help here would be great.
57506
I've seen the identity $\displaystyle \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ used here recently.  I checked for proofs here   I couldn't have figured out the combinatorial proof by myself, and the inductive proof assumes you already know the answer...  So my question is : do you know how to prove directly through computation that $\displaystyle \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ ?
64981
Is there any way to find:  $$\sum_{i=0}^n {\binom{n}{i}i^k}$$  I know that we can find it for small k by using binomial theorem by differentiating both sides and then multiplying both sides by x and repeating till the form presents itself and then putting x =1, but is there a way for any general power( k here) of i?
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
42494
Why does:   $$ \sum_{k=0}^{n} k \binom nk p^k (1-p)^{n-k} = np $$ ? Taking the derivative of:   $$ \sum_{k=0}^{n}  \binom nk p^k (1-p)^{n-k} = (1 + [1-P])^n = 1 $$   does not seem useful, since you would get zero. And induction hasn't yet worked for me, since – during the inductive step – I am unable to prove that:   $$ \sum_{k=0}^{n+1} k \binom {n+1}{k} p^k (1-p)^{n+1-k} = (n+1)p  $$   assuming that:   $$ \sum_{k=0}^{n} k \binom {n}{k} p^k (1-p)^{n-k} = np  $$
18597
Show that $\,\displaystyle\sum_{i=1}^{n} \binom{i}{2}=\binom{n+1}{3}$.   I'm thinking right now (though not getting anywhere with it) that I want to expand out the summation portion to $i!/2!(i-2)!$ and simplify from there? Not sure if that will help, not to mention if I put $1$ in for $i$ I get $1/-2$ which I don't think is right.   Anyone care to shed some light on the subject?   Thanks.
115709
Proving by induction. We'd like to show that $2 + 4 + 6 + \cdots+ 2n = n(n + 1)$.  A nice way to do this is by induction. Let $S(n)$ be the statement above. An inductive proof would have the following steps:  Show that $S(1)$ is true.  Show that if $S(1),\ldots,S(k)$ are true, then so is $S(k + 1)$.    This question is really starting to bug me, am I doing something wrong or is the equation wrong for that series? It only seems to work for $S(1)$ but after that it does not give the correct series. The $n(n+1)$ series is $2, 6, 12,\ldots.$ I'm really confused can someone please nudge me in the right direction (I know how proofs by induction works) I'm just having problems with this one in particular.
10141
Prove: $\binom{n}{0}F_0+\binom{n}{1}F_1+\binom{n}{2}F_2+\cdots+\binom{n}{n}F_n=F_{2n}$;  I was stuck with this question for a while... Help me please!!! Thanks!!!
93890
I try to prove the identity:  $$\sum_{k=0}^n \binom{2n+1}{2k}=2^{2n}$$  but have troubles with the last step:  $$\sum_{k=0}^n \binom{2n+1}{k}=\sum_{k=0}^n  \left( \binom{2n}{2k} + \binom{2n}{2k-1}\right) =\sum_{j=-1}^{2n}\binom{2n}{j}=2^{2n}$$  But should not $j$ start from $0$? Is something wrong?