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\(E=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{101}{3^{101}}\)
\(\Leftrightarrow3E=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{101}{3^{100}}\)
\(\Leftrightarrow3E-E=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{101}{3^{100}}-\frac{1}{3}-\frac{2}{3^2}-\frac{3}{3^3}-...-\frac{101}{3^{101}}\)
\(\Leftrightarrow2E=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{100}{3^{101}}\)
Đặt \(S=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(\Leftrightarrow3S=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(\Leftrightarrow3S-S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{1}{3}-\frac{1}{3^2}-...-\frac{1}{3^{100}}\)
\(\Leftrightarrow2S=1-\frac{1}{3^{100}}\)
\(\Leftrightarrow S=\left(1-\frac{1}{3^{100}}\right)\div2\)
\(\Leftrightarrow2E=1+\left(1-\frac{1}{3^{100}}\right)\div2-\frac{101}{3^{101}}\)
\(\Leftrightarrow2E=1+\frac{1}{2}-\frac{1}{3^{100}.2}-\frac{101}{3^{101}}\)
\(\Leftrightarrow2E=\frac{3}{2}-\frac{1}{3^{100}.2}-\frac{101}{3^{101}}< \frac{3}{2}\)
\(\Leftrightarrow E< \frac{3}{4}\left(đpcm\right)\)
Ta có ; \(A=3+3^2+3^3+.....+3^{100}\)
\(=\left(3+3^2+3^3+3^4+3^5\right)\)
B = 3+32 +...+3100
=> B = (3+32+33+34)+(35+36+37+38)+.....+(397+398+399+3100)
=> B = 120 + 34 . 120 +......+396 . 120
=> B = 120.(1+34+38+....+396) chia hết cho 120
=> B chia hết cho 120
Cho Mình
\(E=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3E=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3E-E=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)
\(2E=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6E=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6E-2E=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4E=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4E=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4E=3-\frac{203}{3^{100}}< 3\)
\(\Rightarrow4E< 3\)
\(\Rightarrow E< \frac{3}{4}\left(đpcm\right)\)
Bài 1:
Ta có: \(3+3^2+3^3+...+3^{100}\)
\(=\left(3+3^2+3^3+3^4\right)+....+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(=120+3^5\left(3+3^2+3^3+3^4\right)+....+3^{96}\left(3+3^2+3^3+3^4\right)\)
\(=120+3^5.120+...+3^{96}.120\)
\(=120.\left(1+3^5+.....+3^{96}\right)\)
\(\Rightarrow3+3^2+3^3+3^4+....+3^{100}\)chia hết cho 120 (vì có chứa thừa số 120)