Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=1-\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{99}+\frac{1}{99}\right)-\frac{1}{100}\)
\(=1+0+0+...+0-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}< 11\)
Vậy : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}< 11\)
=1/2-1/3+1/3-1/4+...+1/99-1/100
=1/2-1/100
=50/100-1/100
=49/100<1
=> dãy trên < 1 đđcm
\(\Leftrightarrow x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\right)=\frac{1}{100}+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow x-\frac{98}{99}=\frac{1}{99}\Leftrightarrow x=1\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
=\(1-\frac{1}{100}=\frac{99}{100}\)
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 = \(\frac{1}{1}\)-\(\frac{1}{100}\)
ai tốt bụng thì tk cho mk nha, mk đg âm điểm đây
A = \(\frac{99}{100}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{98.99}-\frac{1}{99.100}\\
=-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\\
=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\\
=-\left(1-\frac{1}{100}\right)=\frac{-99}{100}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{50}{100}-\frac{2}{100}=\frac{49}{100}\)
~~~~~~~~~~~Ai thấy đúng thì nhanh k nha ~~~~~~
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~ Và chúc các bạn trả lời câu hỏi này kiếm được nhiều k hơn ~~~~~~~~~~~~
\(=\frac{1.2.3.....99}{1.2.3.....98}.\frac{1.2.3......99}{2.3.4.5....100}\)
\(=99.\frac{1}{100}\)
\(=\frac{99}{100}\)
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
Làm bậy, mà đúng
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ … + \(\frac{1}{99.100}\)
= \(\frac{1}{1}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)-\(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+ … + \(\frac{1}{99}\)- \(\frac{1}{100}\)
= \(\frac{1}{1}\)- \(\frac{1}{100}\)
= \(\frac{99}{100}\)
Công thức :\(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)
Áp dụng:
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Vậy.................
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)