Question

3(2²+1)(2⁴+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)=
a.3(2^126+1)
b.1/2(2^126-1)
c.3(2^126-1)
d.2^128-1
e.2^128+6.2^64+1

Answer 1

Ubah 3 menjadi 2² - 1
(2² - 1)(2² + 1)(2⁴ + 1)(2¹⁶ + 1)(2³² + 1)(2⁶⁴ + 1)
= 2¹²⁸ - 1
Jawaban: D

Answer 2

Ingat : 3 = 2^2-1
maka (2^2-1)(2^2+1) = 4^2-1
=> bentuk aljabar (a+b)(a-b) = a^2-b^2
sehingga hasil akhir = (2^64-1)(2^64+1)
= 2^128-1 D