Would this approach be counted as a combinatorial proof? Or is it just a normal proof??
My Approach
From Binomial Theorem
$(x+a)^{n}$ = $\sum_{k=1}^{n} {a^{n-k}}{x^k}{n \choose k}$ Let’s say a=1.
∴$(x+1)^{n}$ = $\sum_{k=1}^{n} {1^{n-k}}{x^k}{n \choose k}$
$(x+1)^{n}$ = $\sum_{k=1}^{n} {x^k}{n \choose k}$
Taking the Derivative w.r.t x on both sides,
$n(x+1)^{n-1}$ = $\sum_{k=1}^{n}k{x^{k-1}} {n \choose k}$
From this, we can say that for all values of x, the above equation is true. Let’s say x=1
∴$n(1+1)^{n-1}$ = $\sum_{k=0}^{n}k {n \choose k}$
∴$n2^{n-1}$ = $\sum_{k=0}^{n}k {n \choose k}$'
Aproach 2
Firstly, we pick a team of k people from the n members available, then choose one of the k members of team to be the Captain. There are ${n \choose k}$ combinations to choose a team of k people, after which there are $k$ options for the captain. As there are different cases . So, we have $\sum_{k=1}^n {n \choose k}k$.
On the other hand, we could first pick one person from the $n$ people to be the captain. Then for each of the remaining unchosen $n-1$ people, they can be either in or out of the team. That's $2^{n-1}$ possibilities. So, we have $n2^{n-1}$.