Show $\sum\limits_{k=1}^n{n-1\choose k-1} =2^{n-1}$ 
*

*Given $$\sum\limits_{k=1}^n k{n\choose k} = n\cdot 2^{n-1}$$





*

*I know that $$k\cdot{n\choose k}=n\cdot{n-1\choose k-1}=(n-k+1)\cdot{n\choose k-1}$$
Therefore $$\sum\limits_{k=1}^n k{n\choose k} = \sum\limits_{k=1}^n n{n-1\choose k-1} = n\cdot 2^{n-1}$$
So, $$n\cdot\sum\limits_{k=1}^n {n-1\choose k-1} = n\cdot 2^{n-1}$$
Therefore $$\sum\limits_{k=1}^n{n-1\choose k-1} =2^{n-1}$$

How is $\quad\sum\limits_{k=1}^n{n-1\choose k-1} =2^{n-1}\quad$?

$$2^{n-1}=(1+1)^{n-1}=\sum_{k=0}^{n-1}\binom{n-1}{k}\cdot 1^{n-1-k}\cdot 1^k=\sum_{k=0}^{n-1}\binom{n-1}{k}=\sum_{k=1}^{n}\binom{n-1}{k-1}.$$
