Prove that $$f(n)=\sum_{k=0}^{n}\binom{2n+k}{n}\binom{n-1+k}{n-1}\equiv1\pmod2,\quad\forall n\in N^{+}$$
I have tried:
By Lucas TH, we have $f(2n)\equiv f(n)\pmod 2$, but $$f(2n+1)=\sum_{k=0}^{n}\binom{2n+1+k}{n}\binom{n+k}{n}\pmod 2.$$ Then, a little hard to induce it on $n$.
Using the hockey-stick identity, write $$ {2n+k \choose n} = {2n+k \choose n+k} =\sum_{i=0}^{n} {n+i+k-1 \choose n+k-1}. $$ Then, $$ \sum_{k=0}^n {2n+k \choose n}{n+k-1 \choose n-1} \\ =\sum_{k=0}^n\sum_{i=0}^n {n+i+k-1 \choose n+k-1}{n+k-1 \choose n-1}\\ = \sum_{k=0}^n\sum_{i=0}^n \frac{(n+i+k-1)!}{(n+k-1)!i!}\cdot \frac{(n+k-1)!}{(n-1)!k!} = \sum_{k=0}^n \sum_{i=0}^n \frac{(n+i+k-1)!}{(n-1)!i!k!}\\ = \sum_{k=0}^n \frac{(n+2k-1)!}{(n-1)!(k!)^2} + 2\sum_{0\le i<k\le n}\frac{(n+i+k-1)!}{(n-1)!i!k!}\\ \equiv \sum_{k=0}^n \frac{(n+2k-1)!}{(n-1)!(k!)^2} = \sum_{k=0}^n {n+2k-1 \choose n-1}{2k \choose k} \pmod 2. $$ For $k\ge 1$, ${2k \choose k}$ is even, so $$ \sum_{k=0}^n {2n+k \choose n}{n+k-1 \choose n-1} \equiv 1 \pmod 2, $$ as required.