Câu hỏi của Biêtdongsaigon - Toán lớp 6 - Học toán với OnlineMath

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Lời giải:
Đặt \(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-....+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

\(3A=1-\frac{2}{3}+\frac{3}{3^2}-.....+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow 4A=A+3A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+....-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(12A=3-1+\frac{1}{3}-\frac{1}{3^2}+...-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

$\Rightarrow 4A+12A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}<3$

$\Rightarrow 16A< 3$

$\Rightarrow A< \frac{3}{16}$

nhưng xl, mk là cn gái ko pải cn trai, muốn ko, thử thj` khắc biết

nhưng mk biết robert lewandoski ở đâu mà đánh

đơn giải như đang giỡn ^ ^

\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...-\frac{100}{3^{100}}\)

\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow3A+A=1+\left(\frac{1}{3}-\frac{2}{3}\right)+\left(\frac{-2}{3^2}+\frac{3}{3^2}\right)+\left(\frac{3}{3^3}-\frac{4}{3^3}\right)+...+\left(\frac{-98}{3^{98}}+\frac{99}{3^{98}}\right)+\left(\frac{99}{3^{99}}-\frac{100}{3^{99}}\right)-\frac{100}{3^{100}}\)

\(\Rightarrow4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow3.4A=3-1+\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow3.4A+4A=3+\left(1-1\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{3^2}-\frac{1}{3^2}\right)+...+\left(\frac{1}{3^{98}}-\frac{1}{3^{98}}\right)-\frac{101}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow16A=3-\frac{99}{3^{99}}-\frac{100}{3^{100}}< 3\Rightarrow A< \frac{3}{16}< \frac{3}{4}\)

a, S= 1.2 + 2.3 + 3.4 + 4.5 + 99.100

   -S= 1/1 - 1/2 + ......... + 1/4 -1/5 + [-(99.100)]

      = 1/1 - 1/5 + [-(99.100)]

      = 4/5 - 99/100

      =-19/100

S  = 19/100

Vậy S = 19/100

k mk nha

a) \(S=1.2+2.3+...+99.100\)

\(\Rightarrow3S=1.2.3+2.3.3+...+99.100.3\)

\(=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)\)

\(=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100\)

\(=99.100.101\)

\(=999900\)

\(\Rightarrow S=\frac{999900}{3}=333300\)

\(A=\frac{\left(1+2+3+...+100\right)\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{2}\right)\left(63.1,2-21.3,6+1\right)}{1-2+3-4+....+99-100}\)

\(=\frac{\frac{100\left(100+1\right)}{2}\left(\frac{3+2-6}{12}\right)\left[63\left(1,2-1,2\right)+1\right]}{\left(1-2\right)+\left(3-4\right)+....+\left(99-100\right)}\)

\(=\frac{5050.\left(-\frac{1}{12}\right).1}{-1+\left(-1\right)+\left(-1\right)+...+\left(-1\right)}\)

\(=\frac{2525.\left(-\frac{1}{6}\right)}{-50}=\frac{101}{12}\)

101/12 bạn nha

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