Tính : `A = 2(3^2 + 1)(3^4 + 1)(3^8 +1)….(3^{64} + 1)`

Đáp án:

$A = \dfrac{3^{128} -1}{4}$

Giải thích các bước giải:

$A= 2(3^2 + 1)(3^4 + 1)(3^8 +1)\dots(3^{64} + 1)$

$\to A = (3^1 – 1)(3^2 + 1)(3^4 + 1)(3^8 +1)\dots(3^{64} + 1)$

$\to 4A = (3^1 – 1)(3^1 + 1)(3^2 + 1)(3^4 + 1)(3^8 +1)\cdots(3^{64} + 1)$

$\to 4A = (3^2 – 1)(3^2 + 1)(3^4 + 1)(3^8 +1)\dots(3^{64} + 1)$

$\to 4A = (3^4 – 1)(3^4 + 1)(3^8 +1)\dots(3^{64} + 1)$

$\to 4A = (3^8 -1)(3^8 +1)\dots(3^{64} + 1)$

$\to 4A = (3^{64} -1)(3^{64} -1)$

$\to 4A = 3^{128} -1$

$\to A = \dfrac{3^{128} -1}{4}$