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Câu hỏi của Trần Minh Hưng - Toán lớp | Học trực tuyến
\(A=1-3+3^2-3^3+...+3^{98}-3^{99}\)
\(\Rightarrow A=\left(1-3+3^2-3^3\right)+...+\left(3^{96}-3^{97}+3^{98}-3^{99}\right)\)
\(\Rightarrow A=\left(1-3+9-27\right)+...+3^{96}.\left(1-3+3^2-3^3\right)\)
\(\Rightarrow A=-20+...+3^{96}.\left(-20\right)\)
\(\Rightarrow A=\left(-20\right).\left(1+...+3^{96}\right)⋮4\)
\(\Rightarrow A⋮4\)
Vậy \(A⋮4\)
A=1-3+32-33+34-35+36-37+...+398-399
=(1-3+32-33)+(34-35+36-37)+...+(396-397+398-399)
=(1-3+32-33)+34(1-3+32-33)+...+396(1-3+32-34
=(1-3+32-33) (1+34+...+396)
=-20 (1+34+...+396):4 vì 20:4
Vậy A:4
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{99}{3^{98}}+\frac{100}{3^{99}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}...+\frac{99}{3^{98}}+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6A-2A=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{203}{3^{100}}< 3\)
\(A< \frac{3}{4}\)
Giải:
a) Ta có:
\(S=1-3+3^2-3^3+...+3^{98}-3^{99}\)
\(=\left(1-3+3^2-3^3\right)+...+\left(3^{96}-3^{97}+3^{98}-3^{99}\right)\)
\(=1\left(1-3+3^2-3^3\right)+...+3^{96}\left(1-3+3^2-3^3\right)\)
\(=1.\left(-20\right)+3^4.\left(-20\right)+...+3^{96}.\left(-20\right)\)
\(=-20.\left(1+3^4+...+3^{96}\right)\)
\(\Rightarrow S⋮-20\) Hay \(S\in B\left(-20\right)\) (Đpcm)
b) Ta có:
\(S=1-3+3^2-3^3+...+3^{98}-3^{99}\)
\(\Rightarrow3S=3-3^2+3^3-3^4+...+3^{99}-3^{100}\)
\(\Rightarrow3S+S=\left(1-3+3^2-3^3+...+3^{98}-3^{99}\right)+\left(3-3^2+3^3-3^4+...+3^{99}-3^{100}\right)\)
\(\Rightarrow4S=1-3^{100}\)
\(\Rightarrow S=\dfrac{1-3^{100}}{4}\)
Mà \(S\in B\left(-20\right)\Rightarrow S\in Z\)
\(\Leftrightarrow1-3^{100}⋮4\) Hay \(3^{100}-1⋮4\Rightarrow3^{100}\div4\) dư \(1\)
Vậy \(3^{100}\) chia cho \(4\) dư \(1\) (Đpcm)
Ta có
\(A=3+3^2+3^3+3^4+3^5+.....+3^{96}+3^{97}+3^{98}+3^{99}+3^{100}\)
\(A=\left(3+3^2+3^3+3^4+3^5\right)+....+\left(3^{96}+3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(A=363+....+3^{95}.363\)
Vì 363⋮121⇒A⋮121