solve $\sum_{k=0}^a\binom{2a}{k}k$

Question

solve $$\sum_{k=0}^a\binom{2a}{k}k$$

I solved $S=\sum_{k=0}^a\binom{2a}{k}$ using $\binom{2a}{k}=\binom{2a}{2a-k}$

and got $S=\frac{4^a+\binom{2a}{a}}{2}$

but this idea doesn't work with $$\sum_{k=0}^a\binom{2a}{k}k$$

is there any Hints or solution for it ?

score 1 · Accepted Answer · answered Sep 19 '14 at 11:29

1

$$\binom{2a}kk=2a\binom{2a-1}{k-1}$$

Now,$$\binom{2a-1}{k-1}=\binom{2a-1}{2a-1-(k-1)}$$

$$\sum_{k=0}^a\binom{2a}kk=2a\sum_{k=1}^{2a}\binom{2a-1}{k-1}=2a\dfrac{\sum_{r=0}^{2a-1}\binom{2a-1}r}2=a(1+1)^{2a-1}$$

answered Sep 19 '14 at 11:29

lab bhattacharjee

1 Answers1