<s>[INST] Prove or disprove ${{2a-1\choose a} + {2a-3\choose a-1} + {2a-5\choose a-2} + \dots {1\choose 1}}={2a\choose a+1}$ Prove or disprove $\displaystyle{{2a-1\choose a} + {2a-3\choose a-1} + {2a-5\choose a-2} + \dots {1\choose 1}}={2a\choose a+1}$
This is not homework. I'm trying to prove something related to Catalan numbers, and I'm stuck here. 
I tried this: this is the coefficient of $x^a$ in $\displaystyle{(1+x)^{2a-1}+x(1+x)^{2a-3}+\dots x^{a-1}(1+x)}$
If we take $(1+x)^{2a-1}$ as the first term, then this is a geometric series of $a$ terms with $\displaystyle{\frac{x}{(1+x)^2}}$ as the common factor. Applying the formula for the summation of a geometric series, we get $$\displaystyle{\frac{(1+x)^{2a-1}[\displaystyle{\frac{x^a}{(1+x)^{2a}}}-1]}{\displaystyle{\frac{x}{(1+x)^2}-1}}}$$
On solving this, we get $$\displaystyle{[(1+x)^{2a+1}-x^a(1+x)](1+x+x^2)^{-1}}$$ $$\displaystyle{=[(1+x)^{2a+1}-x^a(1+x)][1-x(1+x)+x^2 (1+x)^2 -x^3 (1+x)^3\dots]}$$
Finding the coefficient of $x^a$ in this expression seems to be the sum of mutiple expressions again!
EDIT: Potato has shown that this is fase by substituting $a=3$. Could someone then give the general expression of the sum?
Thanks in advance!
 [/INST] This is false. You can see this by testing $a=3$. 
$${ 5 \choose 3} + { 3 \choose 2 } + 1 = {6 \choose 4}-1\neq {6 \choose 4}$$
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