How do I prove this combinatorial identity Show that 

$${2n \choose n} + 3{2n-1 \choose n} + 3^2{2n-2 \choose n} + \cdots + 3^n{n \choose n} \\ = {2n+1 \choose n+1} + 2{2n+1 \choose n+2} + 2^2{2n+1 \choose n+3} + \cdots + 2^n{2n+1 \choose 2n+1}$$

One way that I did it was to use the idea of generating functions.
For the left hand side expression, I can find 2 functions. Consider;
$$f_1 (x) = \frac{1}{(1-3x)} \\ = 1 + 3^1x + 3^2x^2 + 3^3x^3 + \cdots + 3^nx^n + \cdots  \\ f_2(x) = \frac{1}{(1-x)^{n+1}} \\ = {n \choose n} + {n+1 \choose n}x + {n+2 \choose n}x^2 + \cdots + {2n-1 \choose n}x^{n-1} + {2n \choose n}x^n + \cdots + $$
Consider the coefficient of $x^n$ in the expansion of $f_1 (x) . f_2 (x)$. Then the coefficient will be the expression on the left hand side.
Now we further consider 2 functions for the right-hand side expression.
Consider;
$$f_3 (x) = \frac {1}{(1-2x)} \\ = 1 + 2^1x + 2^2x^2 + \cdots + 2^{n-1}x^{n-1} + 2^nx^n + \cdots \\ f_4 (x) = (1+x)^{2n+1} \\= 1 + {2n+1 \choose 1}x + \cdots + {2n+1 \choose n-1}x^{n-1} + {2n+1 \choose n}x^n +\cdots + {2n+1 \choose 0}x^{2n +1} \\ = {2n+1 \choose 2n+1} + {2n+1 \choose 2n}x + {2n+1 \choose 2n-1}x^2 + \cdots + {2n+1 \choose n+2}x^{n-1} + {2n+1 \choose n+1}x^{n} + \\
+ {2n+1 \choose n}x^{n+1} +\cdots + {2n+1 \choose 0}x^{2n +1}$$
Hence the coefficient of $x^n$ is the coefficient of $x^n$ in the expansion of $f_3(x) . f_4(x)$
This is what I managed to do so far. I'm not sure if $f_1(x) .f_2(x) = f_3(x).f_4(x)$. If the two functions are indeed equal, then I can conclude that their coefficient of $x^n$ must be equal, which will immediately answer the question. If they are equal, how do I show that they are?
If the two functions are not equal? How do I proceed to show this question?
Edit: It might not be true that the product of the two functions are equal. I tried substituting $x=0.1, n=1$. Seems like the two values are not equal. How do I proceed with this question?
 A: We seek to show that
$$\sum_{q=0}^n {2n-q\choose n} 3^q
= \sum_{q=0}^n {2n+1\choose n+1+q} 2^q.$$
We have for the LHS
$$\sum_{q=0}^n {2n-q\choose n-q} 3^q
= \sum_{q=0}^n 3^q [z^{n-q}] (1+z)^{2n-q}
\\ = [z^n] (1+z)^{2n} \sum_{q=0}^n 3^q z^q (1+z)^{-q}.$$
The coefficient extractor controls the range and we obtain
$$[z^n] (1+z)^{2n} \sum_{q\ge 0} 3^q z^q (1+z)^{-q}
= [z^n] (1+z)^{2n} \frac{1}{1-3z/(1+z)}
\\ = [z^n] (1+z)^{2n+1} \frac{1}{1-2z}.$$
We could  conclude at this  point by inspection. Continuing  anyway we
get for the RHS
$$\sum_{q=0}^n {2n+1\choose n-q} 2^q
= \sum_{q=0}^n 2^q [z^{n-q}] (1+z)^{2n+1}
\\ = [z^n] (1+z)^{2n+1} \sum_{q=0}^n 2^q z^q.$$
The coefficient extractor once more controls the range and we obtain
$$[z^n] (1+z)^{2n+1} \sum_{q\ge 0} 2^q z^q
= [z^n] (1+z)^{2n+1} \frac{1}{1-2z}.$$
The two generating functions are the same and we have equality for LHS
and RHS.
A: Using your functions, consider 
$$
3^n f_2(\frac13) = 3^n  \frac{1}{(1-\frac13)^{n+1}} = \frac32 (\frac92)^n\\ = {n \choose n}3^n + {n+1 \choose n}3^{n-1} + \cdots + {2n \choose n} + {\color{red}{ {2n +1 \choose n} 3^{-1}+ \cdots}}
$$
and further
$$
2^n f_4 (\frac12) = 2^n (\frac32)^{2n+1} = \frac32 (\frac92)^n \\= {2n+1 \choose 2n+1}2^n + {2n+1 \choose 2n}2^{n-1}  + \cdots +  {2n+1 \choose n+1}  + {\color{red}{ {2n +1 \choose n} 2^{-1}+ \cdots + {2n +1 \choose 0} 2^{-n-1}}}
$$
The two expressions both equal $\frac32 (\frac92)^n$, and  the first $n+1$ many terms represent the  LHS and RHS of the original question. The terms in red are extra terms: once it is established that these terms also equal, the questions is solved. That is, show  that 
$$
 \sum_{k=1}^{\infty}{2n +k \choose n} 3^{-k} = \sum_{k=1}^{n+1}{2n +1 \choose n+1-k} 2^{-k}
$$
A: Here is a combinatorial proof. Both sides of the equation answer the following question:

How many sequences are there of length $2n+1$, with entries in $\{0,1,2\}$,  such that
  
  
*
  
*at least one of the entries is a $2$, and
  
*there are exactly $n$ zeroes to the left of the leftmost $2$?
  

LHS:
Suppose the leftmost $2$ occurs in spot $k+1$. Among the $k$ spots before hand, you must choose $n$ of the entries to be zero. The $2n+1-(k+1)=2n-k$ spots afterward can be anything. There are $\binom{k}n3^{2n-k}$ ways to do this. Then sum over $k$.
RHS:
Suppose there are $j$ entries which are equal to $0$ or $2$. Choose those entries which are equal to $0$ or $2$ in $\binom{2n+1}j$ ways. The leftmost $n$ of these entries must be zero, the $(n+1)^{st}$ entry must be two, then the remaining $j-(n+1)$ entries can be chosen freely among $0$ and $2$. There are $\binom{2n+1}{j}2^{j-(n+1)}$ ways to do this, then sum over $j$.
