\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)

\(A=\frac{1}{4.6}+\frac{1}{10.12}+\frac{1}{18.20}+...+\frac{1}{810.812}\)

.......

~ Chúc học tốt ~ 

Ai ngang qua xin để lại 1 L - I - K - E

\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{27.28.29.30}\)

\(3A=3.\left(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+......+\frac{1}{27.28.29.30}\right)\)

\(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+..........+\frac{3}{27.28.29.30}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+........+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}\)

\(3A=\frac{1}{6}-\frac{1}{24360}\)

\(3A=\frac{1353}{8120}\)

\(A=\frac{1353}{8120}:3\)

\(A=\frac{451}{8120}\)

Ta có: \(\frac{-3}{1.2.3}+\frac{-3}{2.3.4}+\frac{-3}{3.4.5}+...+\frac{-3}{18.19.20}\)

          \(=\frac{-3}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{18.19.20}\right)\)

          \(=\frac{-3}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

            \(=\frac{-3}{2}\left(\frac{1}{2}-\frac{1}{19.20}\right)=\frac{-3}{2}.\frac{189}{380}=\frac{-567}{760}\)

Lời giải:

$2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2014-2012}{2012.2013.2014}$

$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{2012.2013}-\frac{1}{2013.2014}$

$=\frac{1}{1.2}-\frac{1}{2013.2014}< \frac{1}{2}$

$\Rightarrow A< \frac{1}{2}:2$

Hay $A< \frac{1}{4}$

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{1}{k}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Leftrightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2\)

k=2

chuan 100%

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{14.15.16}\)

\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{14.15.16}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{14.15}-\frac{1}{15.16}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{15.16}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{240}\right)\)

\(=\frac{1}{2}.\frac{119}{240}\)

\(=\frac{119}{480}\)

Bài làm:

Ta có:\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{14.15.16}\)

\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{14.15.16}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{14.15}-\frac{1}{15.16}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{15.16}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{240}\right)\)

\(=\frac{1}{2}.\frac{119}{240}=\frac{119}{480}\)