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Bài 1:
\(2^{49}=\left(2^7\right)^7=128^7;5^{21}=\left(5^3\right)^7=125^7\\ Vì:128^7>125^7\Rightarrow2^{49}>5^{21}\)
Bài 2:
\(a,S=1+3+3^2+3^3+...+3^{99}\\ =\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)\\ =40+3^4.40+...+3^{96}.40\\ =40.\left(1+3^4+...+3^{96}\right)⋮40\\ b,S=1+4+4^2+4^3+...+4^{62}\\ =\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{60}.\left(1+4+4^2\right)\\ =21+4^3.21+...+4^{60}.21\\ =21.\left(1+4^3+...+4^{60}\right)⋮21\)
Bài 1 :
\(2^{49}=\left(2^7\right)^7=128^7\)
\(5^{21}=\left(5^3\right)^7=125^7\)
mà \(125^7< 128^7\)
\(\Rightarrow2^{49}>5^{21}\)
Bài 2 :
a) \(S=1+3+3^2+3^3+...3^{99}\)
\(\Rightarrow S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)...+3^{96}\left(1+3+3^2+3^3\right)\)
\(\Rightarrow S=40+40.3^4+...+40.3^{96}\)
\(\Rightarrow S=40\left(1+3^4+...+3^{96}\right)⋮40\)
\(\Rightarrow dpcm\)
b) \(S=1+4+4^2+4^3+...4^{62}\)
\(\Rightarrow S=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...4^{60}\left(1+4+4^2\right)\)
\(\Rightarrow S=21+4^3.21+...4^{60}.21\)
\(\Rightarrow S=21\left(1+4^3+...4^{60}\right)⋮21\)
\(\Rightarrow dpcm\)
\(S=3+3^2+3^3+...+3^{100}\)
\(\Leftrightarrow3S=3^2+3^3+3^4+...+3^{101}\)
\(\Leftrightarrow3S-S=\left(3^2+3^3+3^4+...+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)
\(\Leftrightarrow2S=3^{101}-3\)
\(\Leftrightarrow S=\frac{3^{101}-3}{2}\)
Ta thấy : \(S=\frac{3^{101}-3}{2}=\frac{\left(3^4\right)^{25}.3-3}{2}=\frac{\overline{...1}.3-3}{2}=\frac{\overline{...3}-3}{2}=\frac{\overline{...0}}{2}=\overline{...0}\)
Vậy chữ số cuối cùng của S là 0
\(A=\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{80}\)
\(\Rightarrow A=\left(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{80}\right)-\left(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{40}\right)\)
\(\Rightarrow A=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{80}\right)-\left(2\cdot\dfrac{1}{2}+2\cdot\dfrac{1}{4}+...+2\cdot\dfrac{1}{80}\right)\)
\(\Rightarrow A=\left(\dfrac{1}{1}+\dfrac{1}{2}-2\cdot\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-2\cdot\dfrac{1}{4}+...+\dfrac{1}{80}-2\cdot\dfrac{1}{80}\right)\)
\(\Rightarrow A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{79}-\dfrac{1}{80}\)
\(\Rightarrow A=\left(\dfrac{1}{1}-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{79}-\dfrac{1}{80}\right)\)
\(\Rightarrow A=\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{79}-\dfrac{1}{80}\right)\)
Ta thấy các biểu thức đằng sau phân số \(\dfrac{1}{2}\) đều dương \(\Rightarrow A>\dfrac{1}{2}\)