Combinatorial identity with binomial coefficients

Question

How can I demonstrate the following relationship? $$\sum_{k \geq 0}\binom{n}{k}k=n2^{n-1}$$

In fact, both of them give me the following numeric sequence: $0, 1, 4, 12, 32,\dots$

EDIT: My apologies I was wrong about the right number

Answer 1

• Look at $$(1+x)^{n} = \sum_{k=0}^{n}\binom{n}{k} x^{k}$$ and differentiate

Answer 2

I assume you meant $... = n2^{n-1}$ at the end.

Assume that you want to choose a team of $k \le n$ members and then choose a captain from it. Then the left side will give all possible combinations of choosing the team first and then the captain for $k \le n$.

Another way to count it is to first choose a captian out of the $n$ people and then decide which of the remaining $n-1$ will be included in the team. This will give you the right hand side.

Obviously these two constructions will give you every possible team of $k \le n$ members and captian combination.

Answer 3

Use the identity $$ \binom{n}{k}\binom{k}{1}=\binom{n}{1}\binom{n-1}{k-1};\quad n\geq k\geq 1. $$ Then $$ \sum_{k \geq 0}\binom{n}{k}k =\sum_{k \geq 1}\binom{n}{k}k =\sum_{k \geq 1}n\binom{n-1}{k-1} =n\sum_{k \geq 1}\binom{n-1}{k-1} =n2^{n-1} $$ as the last sum counts the number of subsets of $\{1,2,\dotsc,n-1\}$ which is $2^{n-1}$ (each element is either in a subset or not).

Answer 4

Lets first look at the LEFT SIDE of the equation, to see what we are counting: We have: $$\sum_{k \geq 0}\binom{n}{k}k=\binom{n}{0}0+\binom{n}{1}1+\binom{n}{2}2+\binom{n}{3}3...$$ 1) Lets analyze the first summand, the $\binom{n}{0}0$, here we select "team" with no members(which corresponds to the selection of an empty set from a set of n elements, which is an empty subset, of which there are only 1 in every subset family ), and as such we can only choose 0 captains for this "team". By the product principle this gives us, 1*0=0.

2) Now lets look at the second summand,the $\binom{n}{1}1$,here we choose a 1 man team ( one man army, :D John J.Rambo) from a class of action heroes.This can happen in $\binom{n}{1}$, ways which corresponds to the number of 1-element subsets of $\{1,2,3,...,n\}$. When we have choosen the one man army hero, only he can be his own captain, in exactly 1 way. Hence, by the product principle we have $\binom{n}{1}1$ ways of choosing this 1 man "team", and it's captain.

3) Next, lets look at all the two man teams we can choose from a class of $\{1,2,3,\cdots,n\}$ students. We can choose this duo, exactly in the same number of ways as we can choose a 2-element subset of $\{1,2,3,\cdots,n\}$ , this number is of course, $\binom{n}{2}$. Let's say that the representative team is Harry and Lloyd. We can choose who will be the captain, in exactly two ways, Harry can be the captain or Lloyd can be the captain. We can do this, by the product principle in $\binom{n}{2}2$. Choosing a 2-subset of n elements, then selecting out one or the other.

I think you get the idea by now, we can construct a bijection which would count the same thing on the right side and we can see that we are counting the same thing, just in a little different way.

THE RIGHT SIDE

Here we start with a class of $\{1,2,3,\cdots,n\}$ students, of which we choose a captain first. We can do this exactly in n ways. Now what does $2^{n-1}$ count. It counts all the subsets of a $\{1,2,3,\cdots,n-1\}$ which we join to our captain choice. When we get the captain in we also get every subset of $\{1,2,3,\cdots,n\}$ since it was the only element missing in every subset.By the product principle we can do it in $n2^{n-1}$ ways. This joining is a bijection. So, the equality formula if proven. That's what I think, I may have had some errors, or have not strictly proven that this is a bijection, but I believe the combinatorial argument stands. I hope this helps.