Prove $\{a_n\}$, where $\sum_{k=0}^{n}\binom n k a_ka_{n-k}=a_n^2$ is a geometric progression Let $\{a_n\}$, $n \geqslant 0$, be a sequence of positive real numbers satisfying $\sum_{k=0}^{n}\binom n k a_ka_{n-k}=a_n^2$. Prove that $\{a_n\}$ is a geometric progression. 
I have tried to consider the ratio of two consecutive terms, and apply Pascal's identity, but to no avail, as I do not know how to deal with the changes to the limits of the sum caused by the use of this identity. Are there any good methods for solving the above?
 A: Note that $a_1^2 = \binom{1}{0}a_1a_0 + \binom 11 a_0a_1 = 2a_0a_1$. This makes $a_0 = \frac 12a_1$.
Similarly,$a_2^2 = a_2a_0 + 2a_1^2 + a_0a_2 = 2a_0a_2 + 8a_0^2 \implies a_2  = 4a_0$. 
Therefore, it's enough to prove that $a_n = \frac{a_{n+1}}{2}$, or in other words, that $a_n = 2^n a_0$.
To do this, note that the base case is done, and the induction step is:
$$
a_n^2 = \sum_{k=0}^n \binom nk a_{n-k}a_k = \sum_{k=1}^{n-1} \binom nk 2^{n-k}2^k a_0^2 + 2a_na_0 = a_0^2 \left(\sum_{k=1}^{n-1} \binom{n}{k}2^n\right)  + 2a_na_0 \\ 
= a_0^2 (2^n-2)2^n + 2a_na_0
$$
From here, I leave you to see that $a_n = 2^na_0$. Hence, the induction is complete.
A: By induction we prove that $a_k = b2^k$.
Base: $a_0^2 = \binom{0}{0}a_0^2 = b^2$
IH: $a_n = b2^n$
$$\begin{align}a_{n+1}^2 &= \sum_{k=0}^{n+1}\binom{n+1}{k}a_ka_{n+1-k} \\
&= ba_{n+1} + \sum_{k=1}^{n}\binom{n+1}{k}b2^kb2^{n+1-k} + ba_{n+1} \\
&= 2ba_{n+1} + b^22^{n+1}(2^{n+1}-2)\end{align}$$
$$\implies a_{n+1}^2-2ba_{n+1} = (b2^{n+1})^2 - 2b(b2^{n+1})$$
$a_{n+1} = b2^{n+1}$
A: The questions is:

Let $\{a_n\}$, $n \geqslant 0$, be a sequence of positive real numbers satisfying $\sum_{k=0}^{n}\binom n k a_ka_{n-k}=a_n^2$. Prove that $\{a_n\}$ is a geometric progression.

First, we show that there exists a geometric progression solution.
Assuming that $a_n$ is a geometric progression, then $a_k\,a_{n-k}$ does not depend on $k$. So $$ a_n^2 = \sum_{k=0}^{n}\binom n k \,a_k\,a_{n-k} = a_0\,a_n\sum_{k=0}^{n}\binom n k = a_0\,a_n 2^n.$$ Since $\,a_n>0,\,$ we divide both sides by it to get
$\,a_n = a_02^n$ which is a geometric progression and it satisfies
the original equation.
Next, we show that the solution is unique.
Rewrite the recursion equation as
$$ a_n^2 = \sum_{k=0}^{n}\binom n k \,a_k\,a_{n-k} = 2\,a_0\,a_n
+ b_n $$
for some $\,b_n>0.\,$ Solve the quadratic to get
$\, a_n = a_0 \pm\sqrt{a_0^2+b_n}. \,$ The minus sign is not
possible because $\,a_0 -\sqrt{a_0^2+b_n}<0\,$ but we already
know that $\,a_n>0.\,$ Thus the solution is unique given the
value of $\,a_0.$
We have shown existence and uniqueness. Q.E.D.
