\(S_n=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(2S_n=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)
\(2S_n=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(2S_n=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(S_n=\frac{1}{4}-\frac{1}{2\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)=\frac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\)
Cách của bạn Đỗ Ngọc Hải cũng đúng . Mik có cách khác nè :
\(S_n=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow S_n=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(\Rightarrow S_n=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(\Rightarrow S_n=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(\Rightarrow S_n=\frac{1}{4}-\frac{1}{2\left(n+1\right)\left(n+2\right)}\)
~ Ủng hộ nhé
\(S_n=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow S_n=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{\left(n+2\right)-n}{n.\left(n+1\right).\left(n+2\right)}\)
\(\Rightarrow2S_n=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2S_n=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2S_n=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2S_n=\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2S_n=\frac{\left(n+1\right)\left(n+2\right)}{2\left(n+1\right)\left(n+2\right)}-\frac{2}{2\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2S_n=\frac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}=\frac{n^2+3n}{2\left(n+1\right)\left(n+1\right)}\)
\(\Rightarrow S_n=\frac{n^2+3n}{2\left(n+1\right)\left(n+2\right)}\div2=\frac{n^2+3n}{2.2\left(n+1\right)\left(n+2\right)}=\frac{n^2+3n}{4\left(n+1\right)\left(n+2\right)}\)
Vay ............................................
Công thức :
\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\left(\frac{3}{6}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)
~ Ủng hộ nhé