0

I have to prove: $$ \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n} = 2^n $$ L.H.S. $$ \frac{n!}{0!(n-0)!} + \frac{n!}{1!(n-1)!} + \frac{n!}{2!(n-2)!} + \cdots + \frac{n!}{n!(n-n)!} $$

$$= 1 + \frac{n!}{(n-1)!} + \frac{n!}{2!(n-2)!} + \cdots + 1 $$

$$= \sum_{k=0}^n \frac{n!}{k!(n-k)!} $$

$$= \sum_{k=0}^n \binom{n}{k} $$

The binomial theorem given in my book is: $$ (a + b)^{m+1} = \sum_{k=0}^{m+1} \binom{m+1}{k} a^{m+1-k} b^k $$

Let m+1=n, a=b=1, as we can raise 1 to the power of anything, $$= \sum_{k=0}^{m+1} \binom{m+1}{k} a^{m+1-k} b^k $$

$$= (1 + 1)^n = 2^n $$

This seems to work until I have to prove

$$ 1\binom{n}{1} + 2\binom{n}{2} + 3\binom{n}{3} + \cdots + n\binom{n}{n} = n2^{n-1}$$ L.H.S.

$$= (\sum_{k=1}^n \binom{n}{k})k $$

Let m+1=n, a=b=1,j+1=k. $$=( \sum_{j=0}^{m+1} \binom{m+1}{j} a^{m+1-j} b^j)k $$ $$= ((1 + 1)^n)k = k2^n $$

Which is clearly wrong. But I don't know why. You can raise 1 to the power of anything, and only the variables are changed, else is fixed.

Praying I don't get question banned, I know my doubts are stupid as hell, sorry.

RobPratt
  • 50,938
xoco
  • 89
  • 2
    I'm not sure that $\sum_{k=1}^n k \binom{n}{k} = (\sum_{k=1}^n \binom{n}{k})n$ – Ronald Dec 17 '24 at 09:47
  • Is the question about the first part or the second part? Could you please clarify? In addition, in the second part, I’m not seeing the formulation of the LHS, perhaps you want to multiply by $k$ instead of $n$? – Michael Burr Dec 17 '24 at 09:47
  • @MichaelBurr, yea that was a typo. The question is about the second part, I can't get R.H.S. – xoco Dec 17 '24 at 10:19
  • @Ronald, fixed it, it was a typo – xoco Dec 17 '24 at 10:19
  • 1
    Why did the question get downvoted? – IamRigour Dec 17 '24 at 11:11
  • Another nice solution to the problem could go as follows: The Powerset of a set with $n$ elements has $2^n$ elements. The binomialcoefficient $\binom{n}{k}$ counts all $k$ elment subsets of a set with $n$ elements. Thus the left and right hand side both just count the subsets of a set with $n$ elements. – Womm Dec 17 '24 at 11:58
  • I'm afraid that your fix isn't correct either because you pulled $k$ out of the sum. Basically the erroneous LHS s where your second proof goes wrong, I think. – Ronald Dec 17 '24 at 13:47
  • @Ronald, you're right. I'm trying to find an answer. – xoco Dec 17 '24 at 16:43

2 Answers2

2

Your expression is $\sum_{k=1}^n\left(\binom nkk\right)$, not $\left(\sum_{k=1}^n\binom nk\right)k$. Indeed, the latter doesn't even make sense, since $k$ is only defined inside the sum, and is not constant.

It's true that $\sum_{k=0}^n\left(\binom nkk\right)=\left(\sum_{k=0}^n\binom nk\right)\overline k$, where $\overline k$ is a "weighted average" value of $k$. Now in this instance it's not hard to see that the weighted average of $k$ must be $n/2$, since $\binom nk=\binom n{n-k}$, and this gives $\frac n22^n=n2^{n-1}$.

(Here I've replaced the lower limit by $0$, since it doesn't change the value - it just adds an extra $0$ term - but makes the expression more symmetrical.)

Another way to put this argument is as follows: $$\sum_{k=0}^nk\binom nk+\sum_{k=0}^n(n-k)\binom nk=n\sum_{n=0}^k\binom nk=n2^n,$$ but since $$\sum_{k=0}^nk\binom nk=\sum_{k=0}^n(n-k)\binom nk,$$ they are both equal to $n2^{n-1}$.

Especially Lime
  • 46,692
0

In the sum $$ 1\binom{n}{1} + 2 \binom{n}{2}+\ldots+ n\binom{n}{n} = \sum_{k=1}^n k\binom{n}{k} $$ you cannot take the $k$ out of the summation, as it is the variable you are summing over.

You can see the identity combinatorially as follows: $$ \sum_{k=1}^n \binom{k}{1}\binom{n}{k} $$

Now $\binom{k}{1}\binom{n}{k}$ which counts pairs of $(k,S)$ consisting of a $k$-element subset $S$ and an element $k\in S$ of that subset by first choosing the subset and then choosing an element of it. Thus the sum just counts pairs $(k,S)$ where $k \in S$ and $S\subseteq \{1,2,\ldots,n\}$ is a subset which contains $k$. Counting in the other order, that is, choosing any $k \in \{1,2,\ldots,n\}$ and then some subset $T$ of the remaining $n-1$ elements and setting $S=T\cup \{k\}$ $$ \sum_{k=1}^n \binom{k}{1}\binom{n}{k} =n.2^{n-1} $$

krm2233
  • 7,230