rú gon bt

V

vipboycodon

$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

= $3\dfrac{(2^2+1)(2^2-1)}{2^2-1}(2^4+1)(2^8+1)(2^{16}+1)$

= $(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)$

= $(2^8-1)(2^8+1)(2^{16}+1)$

= $(2^{16}-1)(2^{16}+1)$

= $2^{32}-1$
 
Top Bottom