Question

[Hoàng Thảo Nguyên]

2^x + 2^x+1 + 2^x+2 + 2^x+3 +...+ 2^x+2015 = 2^2019 + 8 giúp mình với

Answer 1

[Lưu Quang Tuấn Tú]

Ta có : $2^x+2^{x+1}+2^{x+2}+...+2^{x+2015}=2^{2019}-8$

$\iff 2^x\left(1+2+2^2+...+2^{2015}\right)=2^{2019}-8$ (1)

Đặt : $A=1+2+2^2+...+2^{2015}$

$\Rightarrow 2A=2+2^2+2^3+...+2^{2016}$

$\Rightarrow 2A-A=\left(2+2^2+2^3+...+2^{2016}\right)-\left(1+2+2^2+...+2^{2015}\right)$

$\Rightarrow A=2^{2016}-1$

Khi đó (1) trở thành :

$2^x\left(2^{2016}-1\right)=2^{2019}-2^3$

$\iff 2^x\left(2^{2016}-1\right)=2^3\left(2^{2016}-1\right)$

$\iff 2^x=2^3\left(2^{2016}-1\ne 0\right)$

$\iff x=3$

Vậy : $x=3$

Answer 2

[Lưu Quang Tuấn Tú]

x = 3 nhé