Find the product of $(1+\frac{1}{5^{2^n}})$ for $n=0$ to $6$

Question

i.e. $(1+\frac{1}{5})(1+\frac{1}{25})(1+\frac{1}{625})...(1+\frac{1}{5^{64}})$

I have no idea or clue to this problem. Is there a general trick/procedure to this kinds of problem?

Answer 1

You can prove something more general:

$$\forall x\in\mathbb{R}\backslash\{1\},\,\forall n\in\mathbb{N},\,\prod_{i=0}^n\left(1+x^{2^i}\right)=\dfrac{x^{2^{n+1}}-1}{x-1}$$ using some methods like induction, juantheron's answer,...

Answer 2

Let $\displaystyle P = \left(1+\frac{1}{5}\right)\cdot \left(1+\frac{1}{5^2}\right)\cdot \left(1+\frac{1}{5^4}\right)...........\left(1+\frac{1}{5^{64}}\right)$

Now Multiply both side by $\displaystyle \left(1-\frac{1}{5}\right)\;,$ We get

$\displaystyle P\cdot \left(1-\frac{1}{5}\right) = \underbrace{\left(1-\frac{1}{5}\right)\cdot \left(1+\frac{1}{5}\right)}\cdot \left(1+\frac{1}{5^2}\right)\cdot \left(1+\frac{1}{5^4}\right)...........\left(1+\frac{1}{5^{64}}\right)$

Now Using $(a-b)\cdot (a+b) = a^2-b^2\;,$ We get

$\displaystyle P\cdot \left(1-\frac{1}{5}\right) = \underbrace{\left(1-\frac{1}{5^2}\right)\cdot \left(1+\frac{1}{5^2}\right)}\cdot \left(1+\frac{1}{5^4}\right)...........\left(1+\frac{1}{5^{64}}\right)$

Now this process is repeated....

So we get $\displaystyle P\cdot \left(1-\frac{1}{5}\right) = \left(1-\frac{1}{5^{128}}\right)$