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\(\frac{1}{1975^2}+\frac{1}{1976^2}+...+\frac{1}{2017^2}< \frac{1}{1974.1975}+\frac{1}{1975.1976}+...+\frac{1}{2016.2017}\)
\(=\frac{1}{1974}-\frac{1}{1975}+\frac{1}{1975}-\frac{1}{1976}+...+\frac{1}{2016}-\frac{1}{2017}=\frac{1}{1974}-\frac{1}{2017}< \frac{1}{1974}\)
Đặt \(S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2S-S=1-\frac{1}{2^{100}}\)
\(\Rightarrow S=1-\frac{1}{2^{100}}< 1\) (đpcm)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}< 1\)
\(\Rightarrow A< 1\)
Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{10^2}< \frac{1}{9.10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow A< 1-\frac{1}{10}\)
\(\Rightarrow A< 1\left(đpcm\right)\)
Vậy \(A< 1\)
Ta có: B = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{8^2}\)
B = \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{8.8}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}\)
B < \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)
B < \(1-\frac{1}{8}\) < 1
Vậy B < 1
Gọi \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow A=1-\frac{1}{8}=\frac{7}{8}\)
Mà \(A=\frac{7}{8}< 1\left(1\right)\)
\(\frac{1}{1.2}>\frac{1}{2^2}\)
\(\frac{1}{2.3}>\frac{1}{3^2}\)
\(...\)
\(\Rightarrow A>B\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)
\(\Rightarrow B< 1\left(đpcm\right)\)
a, M=1/1.2+1/2.3+...+1/49.50
M=1−1/2+1/2−1/3+...+1/49−1/50
M=1−1/50<1
Vậy M<1
\(a,\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}< 1\)
\(=>M< 1\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};\frac{1}{4^2}< \frac{1}{3\cdot4};....;\frac{1}{100^2}< \frac{1}{99\cdot100}\)
\(\Rightarrow A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}< 1\left(đpcm\right)\)
Ta có : \(\frac{1}{2^2}=\frac{1}{4}< \frac{1}{1.2}\)
\(\frac{1}{3^2}=\frac{1}{9}< \frac{1}{2.3}\)
\(\frac{1}{4^2}=\frac{1}{16}< \frac{1}{3.4}\)
....
\(\frac{1}{100^2}=\frac{1}{10000}< \frac{1}{99.100}\)
Suy ra : \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}< 1\)
Vậy ta có đpcm
a)đặt B=1/2.3+1/3.4+...+1/99.100
=1/1.2+1/2.3+1/3.4+...+1/99.100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100<1 (1)
Mà 1<2(2)
A =1/1+1/2.2+1/3.3+...+1/100.100<1-1/2+1/2-1/3+...+1/99-1/100 (3)
từ (1),(2),(3) =>A<2
b,c tự làm
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}< 1\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}< 1\)
\(S=1-\frac{1}{50}< 1\)
\(S=\frac{49}{50}< 1\left(đpcm\right)\)
Cho em mượn acc bang bang
em có làm được bài này ko?