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9:03 AM
I wanted to vote to close this question
Q: Can't find an identy for proving that $ \sum_{k=0}^{i+1} \binom {i+1} k=2^{i+1}$

user132048 $$ \sum_{k=0}^{i+1} \binom {i+1} k$$ I can't find an identity for this summation :( To clarify I'm trying to prove using induction that this sum is equal to $2^{i+1}$, I have my basis and inductive hypothesis done, this is just my inductive step

as a duplicate of
Q: Prove by induction: $2^n = C(n,0) + C(n,1) + \cdots + C(n,n)$

Jason CurtThis is a question I came across in an old midterm and I'm not sure how to do it. Any help is appreciated. $$2^n = C(n,0) + C(n,1) + \cdots + C(n,n).$$ Prove this statement is true for all $n \ge 0$ by induction.

I got the error message: This closure would result in the 'duplicate of' navigation only leading in a circle.
Probably the circle is cause by this question?
Q: Proving a special case of the binomial theorem: $\sum^{k}_{m=0}\binom{k}{m} = 2^k$

user8367I want to know if I can get some help with this proof. I tried, but I just cannot seem to get $2^{k}$. It states that, For $k \in \mathbb{Z}_{\ge 0}$, $$\sum^{k}_{m=0}\binom{k}{m} = 2^k$$ Thank You.

I voted to close a duplicate of:
Q: Algebraic Proof that $\sum\limits_{i=0}^n \binom{n}{i}=2^n$

JSchlatherI'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 $2^n$.

I still think that the first choice would be a little bit better, since they are both specifically about inductive proofs.
3 hours later…
11:44 AM
@MartinSleziak As you surely noticed, 177405 and 27539 were somehow closed as duplicates of each other. 27539 was also closed as a duplicate of 18690. I changed it so that 18690 is the only stated duplicate of 27539.
12:08 PM
@ArthurFischer Thanks for that!
5 hours later…
4:41 PM
We must have plenty of questions asking about Vandermonde identity:
Q: One Binomial Equation $\sum_{i=0}^{z} {n_1 \choose i}{n_2 \choose z-i} = {n_1+n_2 \choose z}$

user2262504I saw one proof using this formula: $$ \sum_{i=0}^{z} {n_1 \choose i}{n_2 \choose z-i} = {n_1+n_2 \choose z}$$ Can anyone help explain it, thank you!

I cast my vote to close as a duplicate. (Probably there are many other possibilities for the choice of the duplicate.)

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