I'd like some help. My first idea was to use induction, but then I get stuck. The base case works just fine, as you'd imagine, and then...
- If $ \sum_{i=0} ^n j {n \choose j} = n 2^{n-1} \rightarrow \sum_{i=0} ^{n+1} j {n+1 \choose j} = (n+1) 2^{n} $ $$ \sum_{i=0} ^{n+1} j {n+1 \choose j} = \sum_{i=0} ^{n} j {n+1 \choose j} + n+1 = \sum_{i=0} ^{n} j (n! / ((n-j)!j!))*(n+1)/(n+1-j) + n+1$$
- And well I can't seem to do anything else.
Another way, also using induction.
$$ \sum_{i=0} ^{n+1} j {n+1 \choose j} = \sum_{i=0} ^{n} j ({n \choose j} - {m \choose j-1}) + m + 1$$
And then I also get stuck.
I know there's another way of proving this without using induction and only with some properties and a couple of tricks I do not see to see.
Any help? Thanks.
Since you wanted a "trick", consider, $$(1+x)^n=\sum_{i=0}^n\binom{n}{i}x^i$$ Differentiating with respect to $x$, $$n(1+x)^{n-1}=\sum_{i=0}^ni\binom{n}{i}x^{i-1}$$ Now substitute $x=1$.