Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+....+\(\frac{1}{100.101}\)=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+....+\(\frac{1}{100}\)-\(\frac{1}{101}\)=1-\(\frac{1}{101}\)=\(\frac{100}{101}\)
b)\(\frac{1}{1.2.3}\)+\(\frac{1}{2.3.4}\)+....+\(\frac{1}{28.29.30}\)=\(\frac{868}{3480}\)=\(\frac{217}{870}\)
c)\(\frac{1}{1.2.3.4}\)+\(\frac{1}{2.3.4.5}\)+....+\(\frac{1}{27.28.29.30}\)=\(\frac{24354}{438480}\)=\(\frac{451}{8120}\)
\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{97.98.99.100}=\frac{1}{3}.\left(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{97.98.99.100}\right)=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{97.98.99}-\frac{1}{98.99.100}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{98.99.100}\right)=\frac{1}{3}.\left(\frac{1}{6}-\frac{1}{970200}\right)=\frac{1}{18}-\frac{1}{6.970200}\)
\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{97.98.99.100}\)
\(=\frac{1}{3}.\left(\frac{3}{1.2.3.4}+ \frac{3}{2.3.4.5}+...+\frac{3}{97.98.99.100}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{97.98.99}-\frac{1}{98.99.100}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{98.99.100}\right)\)
\(=\frac{1}{3}.\frac{161699}{970200}=\frac{161699}{299106000}\)
=1/1.2-1/3.4+1/2.3-1/3.4+...+1/116.117-1/118.119
=1-1/2-1/3+1/4+1/2-1/3-1/3-1/4+...+1/116-1/117-1/118+1/119
=1+1/119=120/119(ko nhầm thì z)
A=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2015.2016.2017}\)
\(\Leftrightarrow\)A=\(\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}-\frac{2}{2017}\)
\(\Leftrightarrow\)A=\(\frac{1}{1}-\frac{1}{2017}\)
\(\Leftrightarrow\)A=\(\frac{2016}{2017}\)
mk quên:Có \(\frac{2016}{2017}< \frac{1}{4}\) \(\Rightarrow\)S<\(\frac{1}{4}\)
