Question

0,5*3^128-(3^64+1)*(3^32+1)*(3^16+1)*(3^8+1)*(3^4+1)*(3^2+1)*(3+1)=

Answer 1

$(3^{64}+1)(3^{32}+1)(3^{16}+1)(3^8+1)(3^4+1)(3^2+1)(3+1)=\\ \\ =\dfrac{(3^{64}+1)(3^{32}+1)(3^{16}+1)(3^8+1)(3^4+1)(3^2+1)(3+1)(3-1)}{3-1}=\\ \\ \\ =\dfrac{(3^{64}+1)(3^{32}+1)(3^{16}+1)(3^8+1)(3^4+1)(3^2+1)(3^2-1)}{2}=\\ \\ =\dfrac{(3^{64}+1)(3^{32}+1)(3^{16}+1)(3^8+1)(3^4+1)(3^4-1)}{2}=\\ \\ =\dfrac{(3^{64}+1)(3^{32}+1)(3^{16}+1)(3^8+1)(3^8-1)}{2}=\\ \\ =\dfrac{(3^{64}+1)(3^{32}+1)(3^{16}+1)(3^{16}-1)}{2}=\dfrac{(3^{64}+1)(3^{32}+1)(3^{32}-1)}{2}=\\ \\ =\dfrac{(3^{64}+1)(3^{64}-1)}{2}=\dfrac{3^{128}-1}{2}$

Окончательно получаем

$0.5\cdot 3^{128}-\left(\dfrac{3^{128}-1}{2}\right)=0.5\cdot3^{128}-0.5\cdot3^{128}+0.5=0.5$

ответ: 0,5.

Answer 2

ответ и решение во вложении