Câu hỏi của Trần Linh Nhi

Question

2^x+2^x+1+2^x+2+...+2^x+2021=2^2026-16

Toru · Accepted Answer

$2^x+2^{x+1}+2^{x+2}+\dots+2^{x+2021}=2^{2026}-16\\Rightarrow 2^x\cdot(1+2+2^2+\dots+2^{2021})=2^4\cdot(2^{2022}-1)\text{ (1) }$

Đặt $A=1+2+2^2+\dots+2^{2021}$

$2A=2+2^2+2^3+\dots+2^{2022}$

$2A-A=\left(2+2^2+2^3+\dots+2^{2022}\right)-\left(1+2+2^2+\dots+2^{2021}\right)$

$A=2^{2022}-1$

Thay $A=2^{2022}-1$ vào (1), ta được:

$2^x\cdot\left(2^{2022}-1\right)=2^4\cdot\left(2^{2022}-1\right)$

$\Rightarrow2^x=2^4\Rightarrow x=4$

Câu trả lời:

$2^x+2^{x+1}+2^{x+2}+\dots+2^{x+2021}=2^{2026}-16\\Rightarrow 2^x\cdot(1+2+2^2+\dots+2^{2021})=2^4\cdot(2^{2022}-1)\text{ (1) }$

Đặt $A=1+2+2^2+\dots+2^{2021}$

$2A=2+2^2+2^3+\dots+2^{2022}$

$2A-A=\left(2+2^2+2^3+\dots+2^{2022}\right)-\left(1+2+2^2+\dots+2^{2021}\right)$

$A=2^{2022}-1$

Thay $A=2^{2022}-1$ vào (1), ta được:

$2^x\cdot\left(2^{2022}-1\right)=2^4\cdot\left(2^{2022}-1\right)$

$\Rightarrow2^x=2^4\Rightarrow x=4$

Coin Hunter · Answer

$2^x+2^{x+1}+2^{x+2}+...+2^{x+2021}=2^{2026}-16$

Đặt A = 2^x+ 2^x+1 + 2^x+2 + ...+ 2^x+2021

2A = 2^x+1 + 2^x+2 + ...+ 2^x+2022

$\Rightarrow$ 2A - A = (2^x+1 + 2^x+2+ ... + 2^x+2022) - (2^x+ 2^x+1 + ... + 2^x+2021)

$\Rightarrow$ A = 2^x+2022- 2^x

$\Rightarrow$ 2^x+2022- 2^x = 2²⁰²⁶ - 16

$\Rightarrow$ 2^x+2022 - 2^x = 2⁴⁺²⁰²² - 2⁴