**Calculate**:

$latex (3+1). (3^2 +1). (3^4 + 1)(3^8 +1)…. (3^{32} +1)

$

Let

$latex x = (3+1). (3^2 +1). (3^4 + 1)(3^8 +1)…. (3^{2n} +1)

$

Or:

$latex displaystyle

x = sum_{n=0}^{n}(3^{2n} + 1) $

Quite messy to expand out:

$latex displaystyle {

sum_{n=0}^{n} (3^{2n})

+

sum_{n=0}^{n}(1)

= ….

}

$

This 14-year-old vienamese student in Berlin – Hyyen Nguyen Thi Minh discovered a smart trick using the identity:

$latex displaystyle { (a -1).(a + 1) = a^{2} – 1}$

or more general,

$latex displaystyle boxed {

(a^{n} -1).(a^{n} + 1) = a^{2n} – 1

}$

He multiplies x by (3-1):

$latex

x. (3-1) = (3-1)(3+1). (3^{2} +1)… (3^{2n} + 1)

$

$latex 2x = (3^{2} -1). (3^{2} +1)…(3^{2n} + 1)

$

$latex 2x = (3^{4} -1).(3^{4} +1) … (3^{2n} + 1)

$

.

.

.

$latex 2x = (3^{4n} -1) $

$latex displaystyle boxed

{

x = (3^{4n}…

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