Question

[Ngô Châu Bảo Oanh]

 tính tổng :B=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)

Answer 1

[Phạm Tuấn Kiệt]

Ta có nhận xét:

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

Áp dụng công thức trên vào bài tập, ta có:

\(\Rightarrow B=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)

\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)

\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)

\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{1482}\right)\)

\(\Rightarrow B=\frac{1}{2}.\frac{370}{741}=\frac{185}{741}\)

Answer 2

[Nguyễn Trần An Thanh]

 \(\Rightarrow B=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\right)\)

\(\Rightarrow B=\frac{1}{2}\left(\frac{3}{1.2.3}-\frac{1}{1.2.3}+\frac{4}{2.3.4}-\frac{2}{2.3.4}+\frac{5}{3.4.5}-\frac{3}{3.4.5}+...+\frac{39}{37.38.39}-\frac{37}{37.38.39}\right)\)

\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)

\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{38.29}\right)\)

\(\Rightarrow B=\frac{1}{2}.\frac{370}{741}=\frac{185}{741}\)