Closed form for a binomial identity another solution

Question

Is this true?

$$\sum_{j=0}^n{j \cdot \displaystyle\binom{j}{r}} =\displaystyle\frac{(n+1)(r+1)-1}{r+2}\displaystyle\binom{n+1}{r+1}$$

Answer 1

Yes, it's true. See formula 4.35 here (p. 26)

We use

$$ \sum_{k=a}^n \sum_{i=a}^k C_{i,k}=\sum_{i=a}^n \sum_{k=i}^n C_{i,k}\tag{1}$$

Then

$$S=\sum_{k=1}^{n} \binom{k}{m}k=\sum_{k=1}^{n} \binom{k}{m}(\sum_{i=1}^k 1)= \sum_{k=1}^{n} \sum_{i=1}^k \binom{k}{m}= \sum_{i=1}^{n} \sum_{k=i}^n \binom{k}{m} \tag{2} $$

But $$\sum_{k=0}^n \binom{k}{c}=\binom{n+1}{c+1} \tag{3}$$ Hence the inner sum in $(2)$ is: $$\sum_{k=i}^n \binom{k}{m}=\sum_{k=0}^n \binom{k}{m}-\sum_{k=0}^{i-1} \binom{k}{m}=\binom{n+1}{m+1}-\binom{i}{m+1} \tag{4}$$

Further, using $(3)$ once more: $$ \sum_{i=1}^n \binom{i}{m+1}=\binom{n+1}{m+2}=\binom{n+1}{m+1}\frac{n-m}{m+2} \tag{5}$$

Then, putting all together:

$$ S=n \binom{n+1}{m+1}-\frac{n-m}{m+2}\binom{n+1}{m+1}=\frac{nm+n+m}{m+2}\binom{n+1}{m+1}$$

Answer 2

Notice:

$$\sum_{n=a}^{m}n\binom{n}{r}=\frac{\frac{\left(m+r+mr\right)\Gamma\left(2+m\right)}{\Gamma\left(1+m-r\right)}-\frac{a\left(a+ar-1\right)\Gamma\left(a\right)}{\Gamma\left(a-r\right)}}{\Gamma\left(3+r\right)}$$

When, $a=0$:

$$\sum_{n=0}^{m}n\binom{n}{r}=\frac{\frac{\left(m+r+mr\right)\Gamma\left(2+m\right)}{\Gamma\left(1+m-r\right)}+\frac{1}{\Gamma\left(-r\right)}}{\Gamma\left(3+r\right)}$$

So, yes you're right!