Question

[scotty]

Tính tổng

B = \(\frac{1}{1.2.3}\)+    \(\frac{1}{2.3.4}\)+    \(\frac{1}{3.4.5}\)+  ... +   \(\frac{1}{48.49.50}\)

Answer 1

[ღ๖ۣۜLinh's ๖ۣۜLinh'sღ] ★we are one★]

\(B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{49.50}\right)\)

Đến đây bạn tự tính nhé

Answer 2

[ʚɞ βùɱ•๖Ćɦíυღ╰❥]

\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)

\(B=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)

\(B=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{49.50}\right)\)

\(B=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{2450}\right)\)

\(B=\frac{1}{2}\cdot\frac{612}{1225}=\frac{306}{1225}\)

Vậy \(B=\frac{306}{1225}\)