Question

(2^2 + 1)(2^4 +1)(2^8 +1)(2^16 + 1)(2^32 + 1)(2^64 + 1) - 1/3 • 2^128

Answer 1

$(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)(2^{64}+1)=$

$=\frac{(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)(2^{64}+1)}{2^2-1}=$

$=\frac{(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)(2^{64}+1)}{3}=\frac{(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)(2^{64}+1)}{3}=$

$=\frac{(2^{16}-1)(2^{16}+1)(2^{32}+1)(2^{64}+1)}{3}=\frac{(2^{32}-1)(2^{32}+1)(2^{64}+1)}{3}=\frac{(2^{64}-1)(2^{64}+1)}{3}=\frac{2^{128}-1}{3}.$

Вывод: все выражение равно $\frac{2^{128}-1}{3}-\frac{2^{128}}{3}=-\frac{1}{3}.$

Ответ: $-\frac{1}{3}$