* Công thức :
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\left(\frac{3}{6}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{20.21.22}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{20.21}-\frac{1}{21.22}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{21.22}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{462}\right)\)
\(=\frac{1}{2}.\left(\frac{231}{462}-\frac{1}{462}\right)\)
\(=\frac{1}{2}.\frac{230}{462}\)
\(=\frac{115}{462}\)
Chúc bạn học tốt !!!
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Câu hỏi tương tự
Chúc bạn học giỏi =/
tổng quát nè bn
\(\frac{1}{n\left(n+k\right)\left(n+2k\right)}+\frac{1}{\left(n+k\right)\left(n+2k\right)\left(n+3k\right)}+\frac{1}{\left(n+2k\right)\left(n+3k\right)\left(n+4k\right)}\)
\(+...+\frac{1}{\left(n+x.k\right)\left(n+x.k+k\right)\left(n+x.k+2k\right)}\)
\(=\frac{\frac{1}{n\left(n+k\right)}-\frac{1}{\left(n+x.k+k\right)\left(n+x.k+2k\right)}}{2k}\)