So sánh M= 1/ 1.2 + 1/ 2. 3 + ..........+ 1/ 49. 50 với 1
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KP
1
17 tháng 4 2019
\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(M=\frac{1}{1}+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{49}-\frac{1}{49}\right)-\frac{1}{50}\)
\(M=\frac{1}{1}-\frac{1}{50}\)
\(M=\frac{49}{50}\)
Vậy M < 1
NN
1
10 tháng 8 2016
2A = 1 + 1/2 + 1/22 + 1/23 + ... + 1/248+ 1/249
2A - A = (1 + 1/2 + 1/22 + 1/23 + ... + 1/248 + 1/249) - (1/2 + 1/22 + 1/23 + 1/24 + ... + 1/249 + 1/250)
A = 1 - 1/250
N
0
2 tháng 9 2020
\(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=2\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(2A=1+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}\)
\(2A-A=A\)
\(=1+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(=1+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{49}}-\frac{1}{2^{50}}\)
\(=1-\frac{1}{2^{50}}< 1\)
\(\Rightarrow A< 1\)
2 tháng 9 2020
\(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=\text{}\text{}1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(A=1-\frac{1}{2^{50}}\)
Vậy \(A\)< 1
PT
23 tháng 3 2017
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{50}=\dfrac{49}{50}\)
HH
23 tháng 3 2017
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{50}=\dfrac{50}{50}-\dfrac{1}{50}=\dfrac{49}{50}\)
T
3 tháng 5 2019
#)Thắc mắc ?
Cho mk hỏi cái ''với 2'' là j bn ? so sánh ak, nếu là so sánh thì mk giải thế này :
#)Giải :
\(M=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{49.50}\)
\(M=2-1+1-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+...+\frac{2}{48}-\frac{2}{49}+\frac{2}{49}-\frac{2}{50}\)
\(M=2-\frac{2}{50}\)
\(M=1\frac{24}{25}=\frac{49}{25}\)
So sánh \(\frac{49}{25}\)với 2
\(2=\frac{2}{1}=\frac{50}{25}\)
Vì \(\frac{49}{25}< \frac{50}{25}\Rightarrow\frac{49}{25}< 2\Rightarrow M< 2\)
#~Will~be~Pens~#
3 tháng 5 2019
\(M=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{49.50}=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)=2\left(1-\frac{1}{50}\right)=2x\frac{49}{50}=\frac{49}{25}=1\frac{24}{25}\)
Vì M=1 24/25
=>M<2
14 tháng 4 2019
ta có : \(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(B=\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+....+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{50}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{25}\right)\)
\(B=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
\(\Rightarrow\)\(B=A\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(M=1-\frac{1}{50}\)
\(\Rightarrow1>M\)
Ta có: 1/1.2+1/2.3+...+1/49.50
= 1-1/2+1/2-1/3+...+1/49-1/50
= 1-1/50
Ta có: 1-1/50 < 1 (luôn luôn đúng)
=> M<1