Question

a) A = 1+\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+..........+\frac{1}{100^2}\)

Chứng minh rằng A<2

b) B =\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+................+\frac{1}{2012^2}\)

Chứng minh rằng \(\frac{1}{2}-\frac{1}{2013}< B< 1\)

Answer 1

a, \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

\(\Rightarrow A< 1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(\Rightarrow A< 1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Rightarrow A< 1+\left(1-\frac{1}{100}\right)\Rightarrow A< 1+1-\frac{1}{100}\Rightarrow A< 2-\frac{1}{100}\Rightarrow A< 2\left(ĐPCM\right)\)

b, \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\)

\(\Rightarrow B< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2011\cdot2012}\)

\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\)

\(\Rightarrow B< 1-\frac{1}{2012}\Rightarrow B< 1\left(1\right)\)

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\)

\(\Rightarrow B>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2012\cdot2013}\)

\(\Rightarrow B>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2012}-\frac{1}{2013}\)

\(\Rightarrow B>\frac{1}{2}-\frac{1}{2013}\Rightarrow\frac{1}{2}-\frac{1}{2013}< B\left(2\right)\)

Từ (1) và (2) => \(\frac{1}{2}-\frac{1}{2013}< B< 1\)

Answer 2

a)A=1+1/2²+1/3²+....+1/100²

      <1+1/1.2+1/2.3+...+1/99.100=1+1-1/2+1/2-1/3+...+1/99-1/100=2-1/100=199/200<2

b)B=1/2²+1/3²+...+1/2012²

     <1/1.2+1/2.3+...+1/2011.2012=1-1/2+1/2-1/3+...+1/2011-1/2012=1-1/2012=2011/2012

     1/2-1/2013=2011/4026<2011/2012<1