How I can show that $\prod_{i=1}^{n}\frac{3^i(x+1)-2^i}{3^i(x+1)-3\cdot2^{i-1}}=\frac{1}{x}-\frac{1}{x}\left(\frac{2}{3}\right)^n+1$?

Question

How I can show that the following equality is true:

$$\prod_{i=1}^{n}\dfrac{3^i(x+1)-2^i}{3^i(x+1)-3\cdot2^{i-1}}=\dfrac{1}{x}-\dfrac{1}{x}\left(\dfrac{2}{3}\right)^n+1\,?$$

Answer 1

Here we have a telescoping product.

We obtain for $n\geq 1$: \begin{align*} \color{blue}{\prod_{j=1}^n\frac{3^j(x+1)-2^j}{3^j(x+1)-3\cdot 2^{j-1}}} &=\frac{1}{3^n}\prod_{j=1}^n\frac{3^j(x+1)-2^j}{3^{j-1}(x+1)-2^{j-1}}\tag{1}\\ &=\frac{1}{3^n}\,\frac{\prod_{j=1}^n \left(3^j(x+1)-2^j\right)}{\prod_{j=1}^{n}\left(3^{j-1}(x+1)-2^{j-1}\right)}\\ &=\frac{1}{3^n}\,\frac{\prod_{j=1}^n \left(3^j(x+1)-2^j\right)}{\prod_{j=0}^{n-1}\left(3^{j}(x+1)-2^{j}\right)}\tag{1}\\ &=\frac{1}{3^n}\,\frac{3^n(x+1)-2^n}{(x+1)-1}\tag{2}\\ &=\frac{3^nx+3^n-2^n}{3^nx}\\ &\,\,\color{blue}{=1+\frac{1}{x}-\left(\frac{2}{3}\right)^n\frac{1}{x}} \end{align*} and the claim follows.

Comment:

• In (1) we factor out $\frac{1}{3^n}$.

• In (2) we shift the index in the product of the denominator by one to start with $j=0$.

• In (3) we use the telescopic property of the product and cancel equal factors in numerator and denominator.

Answer 2

I could prove the equality by induction. @Marco: Thank you very much for your hint.

I attached the sketch of prove (inductive step). The base case $n=1$ is trivially true.

Solution based on induction