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A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
A < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
A < \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
A < 1 - \(\frac{1.}{100}\)
A < \(\frac{99}{100}< \frac{199}{100}\)
=> A < \(\frac{199}{100}\)
b,
S = \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{99}{10^2}\)
S = \(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{9.11}{10.10}\)
S = \(\frac{1.3.2.4.3.5.4.6.5.7...9.11}{2.2.3.3.4.4...10.10}\)
S = \(\frac{1.2.3^2.4^2.5^2...9^2.10.11}{2^2.3^3.4^2...10^2}\)
S = \(\frac{1.11}{2.10}\)
S = \(\frac{11}{20}\)
2 vế bằng nhau
100-(1+1/2+1/3+...+1/100) = 1/2+2/3+3/4+...+99/100
100- 1-1/2-1/3-...-1/100 = 1/2+2/3+3/4+...+99/100
100 = 1 + 1/2 + 1/2 + 1/3 + 2/3 + ... + 1/100 + 99/100 (cùng cộng 2 vế với (- 1-1/2-1/3-...-1/100)
100 = 1 + 1 + 1 + ... + 1 (100 số hạng)
100 = 100
Vậy 100-(1+1/2+1/3+...+1/100) = 1/2+2/3+3/4+...+99/100
a,\(\frac{-3}{1.3}+\frac{-3}{3.5}+....+\frac{-3}{97.99}\)
= -3.\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)\)
=\(\frac{-3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)\)
=\(\frac{-3}{2}\left(1-\frac{1}{99}\right)\)
=\(\frac{-3}{2}.\frac{98}{99}\)
=\(\frac{49}{-33}\)>\(\frac{49}{-20}\)
\(\text{Đặt biểu thức là A:}\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}+\frac{1}{100^2}\)
\(\text{Ta có:}\frac{1}{2^2}=\frac{1}{2\times2}< \frac{1}{1\times2}\)
\(\frac{1}{3^2}=\frac{1}{3\times3}< \frac{1}{2\times3}\)
\(\frac{1}{4^2}=\frac{1}{4\times4}< \frac{1}{3\times4}\)
\(...\)
\(\frac{1}{99^2}=\frac{1}{99\times99}< \frac{1}{98\times99}\)
\(\frac{1}{100^2}=\frac{1}{100\times100}=\frac{1}{99\times100}\)
\(\Rightarrow A< \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\left(đpcm\right)\)
1.
\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}+\frac{1}{2^{100}}\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\left(\frac{1}{2^{100}}+\frac{1}{2^{100}}\right)\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)
cứ làm như vậy ta được :
\(=1+1=2\)
2. Ta có :
\(\frac{2008+2009}{2009+2010}=\frac{2008}{2009+2010}+\frac{2009}{2009+2010}\)
vì \(\frac{2008}{2009}>\frac{2008}{2009+2010}\); \(\frac{2009}{2010}>\frac{2009}{2009+2010}\)
\(\Rightarrow\frac{2008}{2009}+\frac{2009}{2010}>\frac{2008+2009}{2009+2010}\)
Ta có:B=1/2^2+1/3^2+...+1/100^2<1/1*2+1/2*3+...+1/99*100
B<1-1/100<1
Mà A=1
Nên B<A
k cho mình với nha