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Show that $2n choose n + 32n-1 choose n + 3^22n-2 choose n + cdots + 3^nn choose n \ = 2n+1 choose n+1 + 22n+1 choose n+2 + 2^22n+1 choose n+3 + cdots + 2^n2n+1 choose 2n+1$

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) = frac1(1-3x) \ = 1 + 3^1x + 3^2x^2 + 3^3x^3 + cdots + 3^nx^n + cdots \ f_2(x) = frac1(1-x)^n+1 \ = n choose n + n+1 choose nx + n+2 choose nx^2 + cdots + 2n-1 choose nx^n-1 + 2n choose nx^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;

$f_3 (x) = frac 1(1-2x) \ = 1 + 2^1x + 2^2x^2 + cdots + 2^n-1x^n-1 + 2^nx^n + cdots \ f_4 (x) = (1+x)^2n+1 \= 1 + 2n+1 choose 1x + 2n+1 choose 2x^2 + cdots + 2n+1 choose n-1x^n-1 + 2n+1 choose nx^n \ = 2n+1 choose 2n+1 + 2n+1 choose 2nx + 2n+1 choose 2n-1x^2 + cdots + 2n+1 choose n+2x^n-1 + 2n+1 choose n+1$

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?

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 $binomkn3^2n-k$ 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 $binom2n+1j$ 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 $binom2n+1j2^j-(n+1)$ ways to do this, then sum over $j$.

Using your functions, consider
$$
3^n f_2(frac13) = 3^n frac1(1-frac13)^n+1 = frac32 (frac92)^n\ = n choose n3^n + n+1 choose n3^n-1 + cdots + 2n choose n + cdots
$$
and further
$$
2^n f_4 (frac12) = 2^n (frac32)^2n+1 = frac32 (frac92)^n \= 2n+1 choose 2n+12^n + 2n+1 choose 2n2^n-1 + cdots + 2n+1 choose n+1
$$
The two are equal.