Question

chứng minh:\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}< 1\)

Answer 1

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

A = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

A = \(1-\frac{1}{100}\)

A < 1

Answer 2

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}< 1\)

Answer 3

 \(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A=1-\frac{1}{100}\)

do \(\frac{1}{100}\ne0\Rightarrow1-\frac{1}{100}< 1\)

=> A< 1(đpcm)

tk mk nha

Answer 4

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

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

\(\Rightarrow A=1-\frac{1}{100}\)

\(\Rightarrow A=\frac{99}{100}\)

\(\Rightarrow A< 1\)

Answer 5

A= 1/1 - 1/2 + 1/2 - 1/3 +1/3 - 1/4 + ... + 1/99 - 1/100

A= 1/1 + 0 + 0 + ... + 0 -1/100

A= 99/100 <1

Answer 6

Chào bạn, đây là cách giải:

A =​​\(\frac{1}{1}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{6}\)... + \(\frac{1}{98}\)-\(\frac{1}{99}\)+ \(\frac{1}{99}\)-\(\frac{1}{100}\)

 Vì bạn thấy \(\frac{1}{2}\)+ \(\frac{1}{2}\)= 0 , nên

= \(\frac{1}{1}\)- \(\frac{1}{100}\)

= \(\frac{100}{100}\)- \(\frac{1}{100}\)

= \(\frac{99}{100}\)< 1