Cho: M=1^4+1^42+1^43+...+1^4 mũ1000.So sánh M với 1^3
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![](https://rs.olm.vn/images/avt/0.png?1311)
1) Ta có : 7245 - 7243 = 7243.(722 - 1)
7244 - 742 = 742.(722 - 1)
Vì 7243 > 7242
=> 7243.(722 - 1) > 742.(722 - 1)
=> 7245 - 7243 > 7244 - 742
2) Giải
\(M=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{50}}\)
\(4M=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{49}}\)
Lấy 4M trừ M theo vế ta có :
\(4M-M=\left(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{49}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{50}}\right)\)
\(3M=1-\frac{1}{49}\)
\(M=\left(1-\frac{1}{49}\right):3\)
\(=\frac{1}{3}-\frac{1}{147}< \frac{1}{3}\)
Vậy \(M< \frac{1}{3}\left(\text{đpcm}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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) \(S=1+4+4^2+4^3+...4^{62}\)
\(\Rightarrow S=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...4^{60}\left(1+4+4^2\right)\)
\(\Rightarrow S=21+4^3.21+...4^{60}.21\)
\(\Rightarrow S=21\left(1+4^3+...4^{60}\right)⋮21\)
\(\Rightarrow dpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
M = \(\frac{1}{2}\)+ \(\frac{2}{3}\)+ \(\frac{3}{4}\)+\(\frac{4}{5}\)+ \(\frac{5}{6}\)+ \(\frac{6}{7}\)+ \(\frac{7}{8}\)+ \(\frac{8}{9}\)+ \(\frac{9}{10}\)= \(\frac{17819}{2520}\)
Vậy: M > 1
![](https://rs.olm.vn/images/avt/0.png?1311)
Cho 0<m<4 ,so sánh P=(m+1)(m+2)(m+3)(m+4)(m-5)với 1 kết quả là P < 1.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=1-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)\)
Đặt \(N=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\)
\(2N=1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2N-N=1-\dfrac{1}{2^{10}}\)
\(\Rightarrow N=1-\dfrac{1}{2^{10}}\)
\(\Rightarrow M=1-\left(1-\dfrac{1}{2^{10}}\right)=\dfrac{1}{2^{10}}>\dfrac{1}{2^{11}}\)
Vậy \(M>\dfrac{1}{2^{11}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
M = 4000 > 1 = 13
M = \(1^3\)