Some days ago i read this question but the topic was closed. I wanted to know if my alternative proof is correct.
I want to prove that $\sum_{k=0}^{n}{\binom{2n+1}{k}}2^{n-k} = \sum_{k=0}^{n}{\binom{n+k}{k}}3^{n-k}$ by induction on $n$. The case $n=0$ is trivial:
$ {\binom{2\cdot 0+1}{0}}2^{0-0}=1={\binom{0+0}{0}}3^{0-0}$
Suppose that the thesis is true for $n-1$ and we prove that is true for $n$:
$S_n:=\sum_{k=0}^{n}{\binom{n+k}{k}}3^{n-k}$ and $R_n:=\sum_{k=0}^{n}{\binom{2n+1}{k}}2^{n-k}$ for every $n\in \mathbb{N}$
we want use this fondamental fact:
$\binom{a}{b}=\binom{a-1}{b}+\binom{a-1}{b-1}$ for every $a,b\in\mathbb{N}, b>0$
$S_n=\binom{n}{0}3^n+\sum_{k=1}^{n}{[\binom{(n-1)+k}{k}+\binom{n+(k-1)}{k-1}]}3^{n-k}=$
$=\binom{n}{0}3^n+\sum_{k=1}^{n}{\binom{(n-1)+k}{k}}3^{n-k}+\sum_{k=1}^{n}{\binom{n+(k-1)}{k-1}}3^{n-k}=$
$=\binom{2n-1}{n}+3\sum_{k=0}^{n-1}{\binom{(n-1)+k}{k}}3^{(n-1)-k}-\frac{1}{3}\binom{2n}{n}+\frac{1}{3}\sum_{k=1}^{n+1}{\binom{n+(k-1)}{k-1}}3^{n-(k-1)}=$
$=\binom{2n-1}{n}-\frac{1}{3}\binom{2n}{n}+3S_{n-1}+\frac{1}{3}S_n$
than
$S_n=\binom{2n-1}{n}-\frac{1}{3}\binom{2n}{n}+3S_{n-1}+\frac{1}{3}S_n$
and so
$(1-\frac{1}{3})S_n=\binom{2n-1}{n}-\frac{1}{3}\binom{2n}{n}+3S_{n-1}$
so
$2S_n=3\binom{2n-1}{n}-\binom{2n}{n}+9S_{n-1}$
Now we consider
$R_n=\sum_{k=0}^{n}{\binom{2n+1}{k}}2^{n-k}=\binom{2n+1}{0}2^n+\binom{2n+1}{1}2^{n-1}+\sum_{k=2}^{n}{[\binom{2n-1}{k}+2\binom{2n-1}{k-1}+\binom{2n-1}{k-2}]}2^{n-k}=$
$\binom{2n+1}{1}2^{n-1}-\binom{2n-1}{1}2^{n-1}+\sum_{k=0}^{n}{\binom{2n-1}{k}}2^{n-k}+2\sum_{k=2}^{n}{\binom{2n-1}{k-1}}2^{n-k}+\sum_{k=2}^{n}{\binom{2n-1}{k-2}}2^{n-k}=$
$=\binom{2n-1}{n}-\frac{1}{2}\binom{2n-1}{n-1}+2\sum_{k=0}^{n-1}{\binom{2n-1}{k}}2^{(n-1)-k}+2\sum_{k=1}^{n}{\binom{2n-1}{k-1}}2^{(n-1)-(k-1)}+\frac{1}{2}\sum_{k=2}^{n+1}{\binom{2n-1}{k-2}}2^{(n-1)-(k-2)}$
but $2(n-1)+1=2n-1$ and so it is equal to
$=\binom{2n-1}{n}-\frac{1}{2}\binom{2n-1}{n-1}+2R_{n-1}+2R_{n-1}+\frac{1}{2}R_{n-1}$
and so
$2R_n=2\binom{2n-1}{n}-\binom{2n-1}{n-1}+9R_{n-1}=3\binom{2n-1}{n}-[\binom{2n-1}{n}+\binom{2n-1}{n-1}]+9R_{n-1}=$
$3\binom{2n-1}{n}-\binom{2n}{n}+9R_{n-1}$
Now we have the two equation
$2S_n=3\binom{2n-1}{n}-\binom{2n}{n}+9S_{n-1}$
$2R_n=3\binom{2n-1}{n}-\binom{2n}{n}+9R_{n-1}$
but by inductive hypotesize we have that $S_{n-1}=R_{n-1}$ and so
$2S_n=2R_n$ than
$S_n=R_n$