Question

Tính Tổng : S = \(\frac{1}{2}\)+\(\frac{1}{6}\)+\(\frac{1}{12}\)+....+\(\frac{1}{90}\)

Làm ơn ai đó giúp mk nhé. mk đang cần gấp

 

 

Answer 1

\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)

\(S=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)

\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(S=1-\frac{1}{10}\)

\(S=\frac{9}{10}\)

Answer 2

\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)

\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(S=1-\frac{1}{10}\)

\(S=\frac{9}{10}\)

Answer 3

Ta có:

\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)

\(\Rightarrow S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(\Rightarrow S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(\Rightarrow S=\frac{1}{1}-\frac{1}{10}\)

\(\Rightarrow S=\frac{9}{10}\)

Answer 4

Thanks các bạn nha. Kết bạn với mk để giúp đỡ mk về môn toán nhé.

Answer 5

\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\)

\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\)

\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{9}-\frac{1}{10}\)

\(S=1-\frac{1}{10}\)

\(S=\frac{9}{10}\)