\r\nTa có: $\\Large F(x)=\\int 2^x\\mathrm{d}x=\\dfrac{2^x}{\\mathrm{ln}2}+C$.
\r\n\r\nTheo giả thiết $\\Large F(0)=\\dfrac{1}{\\mathrm{ln}2}$ $\\Large \\Leftrightarrow \\dfrac{2^0}{\\mathrm{ln}2}+C=\\dfrac{1}{\\mathrm{ln}2}$ $\\Large \\Leftrightarrow C=0$. Suy ra: $\\Large F(x)=\\dfrac{2^x}{\\mathrm{ln}2}$
\r\n\r\nVậy $\\Large T=F(0)+F(1)+F(2)+...+F(2019)$ $\\Large =\\dfrac{2^0}{\\mathrm{ln}2}+\\dfrac{2^1}{\\mathrm{ln}2}+\\dfrac{2^2}{\\mathrm{ln}2}+...+\\dfrac{2^{2019}}{\\mathrm{ln}2}$
\r\n\r\n$\\Large =\\dfrac{1}{\\mathrm{ln}2}(2^0+2^1+2^2+...+2^{2019})$ $\\Large =\\dfrac{1}{\\mathrm{ln}2}.1.\\dfrac{1-2^{2020}}{1-2}=\\dfrac{2^{2020}-1}{\\mathrm{ln}2}$.
\r\n","url":"https://hoc357.edu.vn/cau-hoi/goi-large-fx-la-mot-nguyen-ham-cua-ham-so-large-fx2x-thoa-v5984","dateCreated":"2022-08-18T19:16:22.626Z","author":{"@type":"Person","name":"Trần Thanh Hùng"}},"suggestedAnswer":[]}}MỤC LỤC
Gọi $\Large F(x)$ là một nguyên hàm của hàm số $\Large f(x)=2^x$, thỏa mãn $\Large F(0)=\dfrac{1}{\mathrm{ln}2}$. Tính giá trị biểu thức $\Large T=F(0)+F(1)+F(2)+...+F(2019)$.
Lời giải chi tiết:
Chọn A
Ta có: $\Large F(x)=\int 2^x\mathrm{d}x=\dfrac{2^x}{\mathrm{ln}2}+C$.
Theo giả thiết $\Large F(0)=\dfrac{1}{\mathrm{ln}2}$ $\Large \Leftrightarrow \dfrac{2^0}{\mathrm{ln}2}+C=\dfrac{1}{\mathrm{ln}2}$ $\Large \Leftrightarrow C=0$. Suy ra: $\Large F(x)=\dfrac{2^x}{\mathrm{ln}2}$
Vậy $\Large T=F(0)+F(1)+F(2)+...+F(2019)$ $\Large =\dfrac{2^0}{\mathrm{ln}2}+\dfrac{2^1}{\mathrm{ln}2}+\dfrac{2^2}{\mathrm{ln}2}+...+\dfrac{2^{2019}}{\mathrm{ln}2}$
$\Large =\dfrac{1}{\mathrm{ln}2}(2^0+2^1+2^2+...+2^{2019})$ $\Large =\dfrac{1}{\mathrm{ln}2}.1.\dfrac{1-2^{2020}}{1-2}=\dfrac{2^{2020}-1}{\mathrm{ln}2}$.
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