Đặt \(A=2^x+2^{x+1}+...+2^{x+2021}=2^{x+2026-16}\)
Đặt \(2A=2^{x+1}+2^{x+2}+...+2^{x+2022}=2^{x+2027+32}\)
Ta lấy \(2A-A=2^{x+2022}-2^x=2^{2026-16}\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
\(2VT=2^{x+1}+2^{x+2}+2^{x+3}+...+2^{x+2022}\)
\(VT=2VT-VT=2^{x+2022}-2^x\)
\(\Rightarrow2^{x+2022}-2^x=2^{2026}-16\)
\(\Leftrightarrow2^{2022}.2^x-2^x=2^{2026}-2^4\)
\(\Leftrightarrow2^x\left(2^{2022}-1\right)=2^4\left(2^{2022}-1\right)\)
\(\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(2^x+2^{x+1}+2^{x+2}+...+2^{x+2021}=2^{2026}-16\)
\(\Rightarrow2^x\left(1+2^x+2^{x+1}+...+2^{x+2020}\right)=2^{2026}-2^4\)
\(\Rightarrow2^x.\dfrac{2^{x+2020+1}-1}{2-1}=2^4.\left(2^{2022}-1\right)\)
\(\Rightarrow2^x.\left(2^{x+2021}-1\right)=2^4.\left(2^{2022}-1\right)\)
\(\Rightarrow x\in\varnothing\)
Bổ sung giải thích dòng kế cuối
\(...\Rightarrow\left\{{}\begin{matrix}x=4\\x+2021=2022\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=4\\4+2021=2022\left(vô,lý\right)\end{matrix}\right.\)
\(\Rightarrow x\in\varnothing\)
