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\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}< 2\left(đpcm\right)\)
a) \(\frac{7}{5}.\frac{-31}{125}.\frac{1}{2}.\frac{10}{17}.\frac{-1}{2^3}=\frac{7.\left(-31\right).1.10.\left(-1\right)}{5.2.125.17.2^3}=\frac{31.7}{17.125.2^3}=\frac{217}{17000}\)
b) \(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).0=0\)
c) \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{3.4.5...100}{2.3.4...99}=\frac{100}{2}=50\)
d) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-99}{100}=\frac{-\left(1.2.3..99\right)}{2.3.4...100}=-\frac{1}{100}\)
e) \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}=\frac{\left(1.2.3..29\right).\left(3.4.5...31\right)}{\left(2.3.4...30\right).\left(2.3.4...30\right)}\)
\(=\frac{1.31}{30.2}=\frac{31}{60}\)
ĐẶT A= 1/3^2 + 1/4^2 + 1/ 5^2 +......+1/100
TA CÓ:
A=1/3^2 + 1/4^2 +1/5^2 +.....+1/100
=>A=1/3^2 + 1/4^2 +1/5^2 +.....+1/10^2
TA THẤY : 1/3^2 < 1/2.3 ;1/4^2 < 1/3.4;1/5^2 < 1/4.5;....;1/10^2<1/9.10
=>1/3^2 + 1/4^2 +1/5^2 +.....+1/10^2<1/2.3 + 1/3.4 + 1/4.5 +.... +1/9.10
=>A<1/2-1/3 + 1/3 -1/4 + 1/4 - 1/5 +...+1/9-1/10
=>A<1/2-1/10<1/2
=>1/3^2 + 1/4^2 +1/5^2 +.....+1/10^2<1/2
=>1/3^2 + 1/4^2 +1/5^2 +.....+1/100<1/2(ĐPCM)
VẬY 1/3^2 + 1/4^2 +1/5^2 +.....+1/100<1/2
NHỚ K CHO MÌNH NHÉ
\(\frac{1}{100^2}\)or\(\frac{1}{100}\) vậy hả bạn