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LP
7
26 tháng 8 2016
A = 1/1×2 + 1/2×3 + 1/3×4 + .. + 1/99×100
A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
A = 1 - 1/100 < 1
PS
26 tháng 8 2016
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=1-\frac{1}{100}< 1\)
=> ĐPCM
JB
1
24 tháng 2 2017
ta có \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).......\left(\frac{1}{10}-1\right)\)
\(A=-\left(\frac{1}{2}.\frac{2}{3}.....\frac{9}{10}\right)\)
\(A=-\frac{1}{10}\)
vi\(-\frac{1}{10}>-\frac{1}{9}\)
do đó A>\(\frac{-1}{9}\)
NT
0
29 tháng 6 2017
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{18}{19}.\frac{19}{20}\)
\(A=\frac{1}{20}\)
29 tháng 6 2017
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)........\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(\Leftrightarrow A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...........\frac{18}{19}.\frac{19}{20}\)
\(\Leftrightarrow A=\frac{1}{20}>\frac{1}{21}\)
\(\Leftrightarrow A>\frac{1}{21}\)
\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)................\left(1-\frac{1}{100}\right)\)
\(\Leftrightarrow B=\frac{3}{4}.\frac{8}{9}..................\frac{99}{100}\)
\(B=\frac{1.3}{2^2}.\frac{2.4}{3^2}................\frac{9.11}{50^2}\)
\(B=\frac{11}{50}< \frac{11}{21}\)
LV
0
ND
1
12 tháng 4 2016
Ta có: \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(A=\frac{-3}{2^2}.\frac{-8}{3^2}.\frac{-15}{4^2}...\frac{-9900}{100^2}\)
\(A=\frac{\left(-1\right).3}{2^2}.\frac{\left(-2\right).4}{3^2}.\frac{\left(-3\right).5}{4^2}...\frac{\left(-99\right).101}{100^2}\)
\(A=\cdot\frac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-99\right)}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}\)
\(A=\left(-\frac{1}{100}\right).\frac{101}{2}\)
\(A=-\frac{101}{200}\)
L
0
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(A=\left(\frac{2}{2}-\frac{1}{2}\right)\left(\frac{3}{3}-\frac{1}{3}\right)...\left(\frac{19}{19}-\frac{1}{19}\right)\left(\frac{20}{20}-\frac{1}{20}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{18}{19}.\frac{19}{20}\)
\(A=\frac{1.2.3...18.19}{2.3.4...19.20}\)
\(A=\frac{1}{20}\Leftrightarrow A>\frac{1}{21}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}......\frac{19}{20}=\frac{1}{20}>\frac{1}{21}\)
\(\text{Vậy: A lớn hơn 1/21}\)