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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.

joaquinlpereyra
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  • Hint : $\binom{n}{j} = \binom{n}{n-j}$. – achille hui Mar 02 '16 at 16:23
  • Another relevant link: http://math.stackexchange.com/questions/351289/simplify-sum-i-0n-i1-binom-ni – Viktor Vaughn Mar 02 '16 at 16:26
  • Questions such as this one find usually a quick answer on the internet with evident keywords like "combinatorial identity". One reference that I have found in a couple of minutes is this one (page 3) http://math.ucsd.edu/~jverstra/bijections.pdf – Jean Marie Mar 02 '16 at 16:39

1 Answers1

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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 $x=1$.

GoodDeeds
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