Prove $\{a_n\}$, where $\sum_{k=0}^{n}\binom n k a_ka_{n-k}=a_n^2$ is a geometric progression

Question

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?

Answer 1

By induction we prove that $a_k = b2^k$.

Base: $a_0^2 = \binom{0}{0}a_0^2 = b^2$

IH: $a_n = b2^n$

$$\begin{align}a_{n+1}^2 &= \sum_{k=0}^{n+1}\binom{n+1}{k}a_ka_{n+1-k} \\ &= ba_{n+1} + \sum_{k=1}^{n}\binom{n+1}{k}b2^kb2^{n+1-k} + ba_{n+1} \\ &= 2ba_{n+1} + b^22^{n+1}(2^{n+1}-2)\end{align}$$

$$\implies a_{n+1}^2-2ba_{n+1} = (b2^{n+1})^2 - 2b(b2^{n+1})$$

$a_{n+1} = b2^{n+1}$

Answer 2

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.

Answer 3

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.