M=1/3+2/3^2+3/3^3+...+2021 /3^2021
Những câu hỏi liên quan
chứng tỏ rằng: M = 1/3 + 2/3^2 + 3/3^3 + ... + 2021/3^2021 không có giá trị nguyên
chứng tỏ rằng: M = 1/3 + 2/3^2 + 3/3^3 + ... + 2021/3^2021 không có giá trị nguyên
chứng tỏ rằng: M = 1/3 + 2/3^2 + 3/3^3 + ... + 2021/3^2021 không có giá trị nguyên
chứng tỏ rằng: M = 1/3 + 2/3^2 + 3/3^3 + ... + 2021/3^2021 không có giá trị nguyên
chứng tỏ rằng: M = 1/3 + 2/3^2 + 3/3^3 + ... + 2021/3^2021 không có giá trị nguyên
chứng tỏ rằng: M = 1/3 + 2/3^2 + 3/3^3 + ... + 2021/3^2021 không có giá trị nguyên
các anh chị giúp em câu này ạ
chứng tỏ rằng: M = 1/3 + 2/3^2 + 3/3^3 + ... + 2021/3^2021 không có giá trị nguyên
Chứng minh rằng :
\(2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\)
\(2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)\)
\(=\left(1-1\right)+\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{2021}\right)\)
\(=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\)
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Giải:
\(2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\)
Ta có:
\(2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)\)
\(=\left(1-1\right)+\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{2021}\right)\)
\(=0+\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{2020}{2021}\)
\(=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{2020}{2021}\)
Mà \(\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{2020}{2021}=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\)
\(\Rightarrow2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\left(đpcm\right)\)
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\(\dfrac{2}{3}\).\(\dfrac{2022}{2021}\)-\(\dfrac{2}{3}\).\(\dfrac{1}{2021}\)+\(\dfrac{1}{3}\)
\(\dfrac{2}{3}.\dfrac{2022}{2021}-\dfrac{2}{3}.\dfrac{1}{2021}+\dfrac{1}{3}\)
\(=\dfrac{2}{3}.\left(\dfrac{2022}{2021}-\dfrac{1}{2021}\right)+\dfrac{1}{3}\)
\(=\dfrac{2}{3}+\dfrac{1}{3}\)
\(=1\)
\(#Nzgoca\)
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