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A= 1/1-1/2+1/2-1/3+1/4-1/5+...+1/101-1/102
A=1-1/102=102/102-1/102=101/102
ý b thì chờ mình tí tìm cách lập luận đã nhé
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=1-\frac{1}{102}\)
\(A=\frac{101}{102}\)
\(A=\frac{1}{2.2}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(A=\frac{1}{4}+\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(A=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
( gạch bỏ các phân số giống nhau)
\(A=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(A=\frac{1}{4}+\frac{2}{9}\)
\(A=\frac{17}{36}\)
phần b, c bn lm tương tự như phần a nha
b) \(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot...\cdot\frac{100^2}{100\cdot101}=\frac{\left(1\cdot2\cdot3\cdot...\cdot100\right)}{1\cdot2\cdot3\cdot4\cdot...\cdot100}\cdot\frac{\left(1\cdot2\cdot3\cdot...\cdot100\right)}{2\cdot3\cdot4\cdot...\cdot101}=1\cdot\frac{1}{101}=\frac{1}{101}\)
a không biết
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}\)
\(=\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{3.4}-\frac{3}{3.4}+\frac{5}{4.5}-\frac{4}{4.5}+...+\frac{99}{98.99}-\frac{98}{98.99}+\frac{100}{99.100}-\frac{99}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\)
\(=\frac{1}{3}.\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{27.80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\frac{3}{80}\)
\(=\frac{1}{80}< \frac{1}{9}\)
Đặt A=\(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}\)
Ta có:\(\frac{1}{4^2}< \frac{1}{3\cdot4}=\frac{1}{3}-\frac{1}{4}\)
\(\frac{1}{5^2}< \frac{1}{4\cdot5}=\frac{1}{4}-\frac{1}{5}\)
.............................
\(\frac{1}{2011^2}< \frac{1}{2010\cdot2011}=\frac{1}{2010}-\frac{1}{2011}\)
\(\Rightarrow A< \frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\cdot\cdot\cdot+\frac{1}{2010}-\frac{1}{2011}\)
\(=\frac{1}{3}-\frac{1}{2011}< \frac{1}{3}\)
Vậy A<\(\frac{1}{3}\)hay \(\frac{1}{4^2}+\frac{1}{5^2}+\cdot\cdot\cdot+\frac{1}{2011^2}< \frac{1}{3}\)
\(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}< \frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2010\cdot2011}\)
Gọi \(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2010\cdot2011}\)là \(S\)
Ta có:
\(S=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2010\cdot2011}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(=\frac{1}{3}-\frac{1}{2011}< \frac{1}{3}\)
Vì \(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}< S\)mà \(S< \frac{1}{3}\)\(\Rightarrow\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}< \frac{1}{3}\)