I'm self studying Burton's number theory and came across the following problem:
Prove that:
$\binom{n}{1}+2\binom{n}{2}+\cdots+n\binom{n}{n}=n2^{n-1}$
So I tried expanding $n(1+b)^{n-1}$ with the binomial theorem and then let $b=1$, but couldn't really get anywhere. Could someone help?
Thanks
By the Binomial Theorem, $\;\;\displaystyle (1+x)^n=\sum_{k=0}^n \binom{n}{k} x^k$, so differentiating gives
$\;\;\displaystyle n(1+x)^{n-1}=\sum_{k=1}^{n} k\binom{n}{k}x^{k-1}$, and letting $x=1$ gives this result.
Another way to show this is to consider selecting a committee of k people from a group of n people, where $1\le k\le n$, and then selecting a chair of the committee.
The left side counts the number of ways to do this, and the right side counts the number of ways to select the chair first and then choose the other committee members, since there are n choices for the chair and then the rest of the committee is any subset of the remaining n-1 people.
