Evaluating $ \prod_{i=1}^n 3^{2^i} + \sum_{i=1}^n 3^{2^{i+1}-1} \prod_{j=i+1}^n 3^{2^j}$

Question

I'm working on difference equations and got stuck on the algebra with this one:

$$ u_n = \prod_{i=1}^n 3^{2^i} + \sum_{i=1}^n 3^{2^{i+1}-1} \prod_{j=i+1}^n 3^{2^j}$$

I need to simplify this to just a function of n. Any help would be much appreciated.

Answer 1

We obtain \begin{align*} \color{blue}{\prod_{i=1}^n3^{2^i}}&\color{blue}{+\sum_{i=1}^n3^{2^{i+1}-1}\prod_{j=i+1}^n3^{2^j}}\\ &=3^{2^1+2^2+\cdots+2^n}+\sum_{i=1}^n3^{2^{i+1}-1}\cdot 3^{2^{i+1}+2^{i+2}+\cdots+2^{n}}\\ &=3^{2\left(2^n-1\right)}+\sum_{i=1}^n3^{2^{i+1}-1}\cdot 3^{2^{i+1}\left(2^{n-i}-1\right)}\\ &=3^{2\left(2^n-1\right)}+\sum_{i=1}^n3^{2^{i+1}-1}\cdot 3^{2^{n+1}-2^{i+1}}\\ &=3^{2\left(2^n-1\right)}+3^{2^{n+1}-1}\sum_{i=1}^n1\\ &=\frac{1}{9}3^{2^{n+1}}+\frac{1}{3}3^{2^{n+1}}n\\ &\color{blue}{=(3n+1)3^{2^{n+1}-2}} \end{align*}

Answer 2

Hint. One may use that $$ \prod_{i=1}^n 3^{2^i}=3^{\large \sum_{i=1}^n2^i},\qquad \prod_{j=i+1}^n 3^{2^j}=3^{\large \sum_{j=i+1}^n2^j}. $$ Can you take it from here?

Answer 3

$$ \begin{align} &\overbrace{\ \prod_{i=1}^n3^{2^i}\ }^{3^{2^{n+1}-2}}+\sum_{i=1}^n3^{2^{i+1}-1}\overbrace{\ \prod_{j=i+1}^n3^{2^j}\ }^{3^{2^{n+1}-2^{i+1}}}\\ &=3^{2^{n+1}-2}+\sum_{i=1}^n3\cdot3^{2^{n+1}-2}\\[6pt] &=(3n+1)3^{2^{n+1}-2} \end{align} $$