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\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
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Lời giải:
Đặt biểu thức là $A$
\(A=\frac{1}{3}+\frac{1}{3^3}+\frac{1}{3^5}+....+\frac{1}{3^{99}}+\frac{1}{3^{101}}\)
\(3^2.A=3+\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
Trừ theo vế:
\(8A=3-\frac{1}{3^{101}}\Rightarrow A=\frac{3}{8}-\frac{1}{8.3^{101}}\)
Akai Haruma Giáo viên Giúp em câu em gửi trong inb nhé chị
P/s : Sorry bạn chủ tus nhé , mình lượn ngay đây
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\(A=3^{100}-3^{99}+3^{98}-...-3+1\\ \Rightarrow\dfrac{1}{3}A=3^{99}-3^{98}+3^{97}-...-1+\dfrac{1}{3}\\ \Rightarrow\dfrac{4}{3}A=3^{100}+\dfrac{1}{3}\\ \Rightarrow A=\dfrac{3^{101}}{4}+\dfrac{1}{4}\)
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a: \(A=3^{100}-3^{99}+3^{98}-...+3^2-3\)
=>\(3A=3^{101}-3^{100}+3^{99}-...+3^3-3^2\)
=>\(4A=3^{101}-3\)
=>\(A=\dfrac{3^{101}-3}{4}\)
b: \(B=\left(-2\right)^0+\left(-2\right)^1+...+\left(-2\right)^{2024}\)
=>\(B\cdot\left(-2\right)=\left(-2\right)^1+\left(-2\right)^2+...+\left(-2\right)^{2025}\)
=>\(-2B-B=\left(-2\right)^1+\left(-2\right)^2+...+\left(-2\right)^{2025}-\left(-2\right)^0-\left(-2\right)^1-...-\left(-2\right)^{2024}\)
=>\(-3B=-2^{2025}-1\)
=>\(B=\dfrac{2^{2025}+1}{3}\)
c: \(C=\left(-\dfrac{1}{5}\right)^0+\left(-\dfrac{1}{5}\right)^1+...+\left(-\dfrac{1}{5}\right)^{2023}\)
=>\(\left(-\dfrac{1}{5}\right)\cdot C=\left(-\dfrac{1}{5}\right)^1+\left(-\dfrac{1}{5}\right)^2+...+\left(-\dfrac{1}{5}\right)^{2024}\)
=>\(\left(-\dfrac{6}{5}\right)\cdot C=\left(-\dfrac{1}{5}\right)^{2024}-\left(-\dfrac{1}{5}\right)^0\)
=>\(C\cdot\dfrac{-6}{5}=\dfrac{1}{5^{2024}}-1=\dfrac{1-5^{2024}}{5^{2024}}\)
=>\(C\cdot\dfrac{6}{5}=\dfrac{5^{2024}-1}{5^{2024}}\)
=>\(C=\dfrac{5^{2024}-1}{5^{2024}}:\dfrac{6}{5}=\dfrac{5^{2024}-1}{6\cdot5^{2023}}\)
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1)
a)\(0,\left(32\right)+0,\left(67\right)\)
\(=0,\left(01\right).32+0,\left(01\right).67\)
\(=0,\left(01\right).\left(32+67\right)\)
\(=\frac{1}{99}.99\)
\(=1\left(đpcm\right)\)
b)\(0,\left(33\right).3\)
\(=0,\left(01\right).33.3\)
\(=\frac{1}{99}.33.3\)
\(=\frac{33}{99}.3\)
\(=\frac{99}{99}\)
\(=1\left(đpcm\right)\)
2)\(0,\left(12\right):1,\left(6\right)=x:0,\left(3\right)\)
\(\left[\left(0,01\right).12\right]:\left[1+0,\left(6\right)\right]=x:\left[0,\left(1\right).3\right]\)
\(\left(\frac{1}{99}.12\right):\left[1+0,\left(1\right).6\right]=x:\left(\frac{1}{9}.3\right)\)
\(\frac{4}{33}:\left[1+\frac{1}{9}.6\right]=x:\frac{1}{3}\)
\(\frac{4}{33}:\left[1+\frac{2}{3}\right]=x.3\)
\(3x=\frac{4}{33}:\frac{5}{3}\)
\(3x=\frac{4}{33}\cdot\frac{3}{5}\)
\(3x=\frac{4}{55}\)
\(x=\frac{4}{55}:3\)
\(x=\frac{4}{55}\cdot\frac{1}{3}\)
\(x=\frac{4}{165}\)
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\(A=\dfrac{1}{2}+\dfrac{1}{301}+\dfrac{1}{302}+...+\dfrac{1}{400}\)
Do \(\dfrac{1}{301}< \dfrac{1}{300};\dfrac{1}{302}< \dfrac{1}{300};...;\dfrac{1}{400}< \dfrac{1}{300}\)
\(\Rightarrow A< \dfrac{1}{2}+\dfrac{1}{300}+\dfrac{1}{300}+...+\dfrac{1}{300}+\dfrac{1}{300}\) (100 số \(\dfrac{1}{300}\) )
\(\Rightarrow A< \dfrac{1}{2}+\dfrac{100}{300}\)
\(\Rightarrow A< \dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}< 1\)
Vậy \(A< 1\)
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Bài 1:
13 + 23 = 1 + 8 = 9 = 32 (là một số chính phương)
13 + 23 + 33 = 1 + 8 + 27 = 36 = 62 (là một số chính phương)
13 + 23 + 33 + 43 = 1 + 8 + 27 + 64 = 100 = 102 (là số cp)
13 + 23 + 33 + 43 + 53 = 1 + 8 + 27 + 64 + 125 = 225 = (15)2 là số cp
Bài 2:
1262 + 1 = \(\overline{..6}\) + 1 = \(\overline{...7}\) (không phải số chính phương)
100! + 8 = \(\overline{...0}\) + 8 = \(\overline{...8}\) (không phải là số chính phương)
1012 - 3 \(\overline{..01}\) - 3 = \(\overline{...8}\) (không phải là số chính phương)
107 + 7 = \(\overline{..0}\) + 7 = \(\overline{..7}\) (không phải là số chính phương)
11 + 112 + 113 = \(\overline{..1}\)+ \(\overline{..1}\)+ \(\overline{..1}\) = \(\overline{...3}\) (không phải số chính phương)