Question

[Nguyễn Hoàng Anh]

Tính :

a) \(\frac{2}{6}\) + \(\frac{2}{12}\)+ \(\frac{2}{20}\)+ \(\frac{2}{30}\)+...+ \(\frac{2}{90}\)+ \(\frac{2}{110}\)

b) \(1\)- \(\frac{1}{2}\)- \(\frac{1}{4}\)-\(\frac{1}{8}\)- \(\frac{1}{16}\)- \(\frac{1}{32}\)- .........- \(\frac{1}{1024}\)- \(\frac{1}{2048}\)

Cảm ơn nhiều ^^

Answer 1

[Xyz OLM]

a) Ta có\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{110}=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{10.11}\)

\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\right)=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{11}\right)=1-\frac{2}{11}=\frac{9}{11}\)

b) Ta có \(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{2048}=1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)(1)

Đặt S = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}+\frac{1}{2048}\)

=> \(2S=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)

Lấy 2S trừ S ta có :

2S - S \(=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}+\frac{1}{2048}\right)\)

\(S=1-\frac{1}{2048}\)

Khi đó (1) <=> \(1-\left(1-\frac{1}{2048}\right)=1-1+\frac{1}{2048}=\frac{1}{2048}\)

Answer 2

[Trang]

\(a,\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+....+\frac{2}{90}+\frac{2}{110}\)

\(=2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+.....+\frac{1}{90}+\frac{1}{110}\right)\)

\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{9.10}+\frac{1}{10.11}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-....+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{11}\right)\)

\(=1-\frac{2}{11}\)

\(=\frac{9}{11}\)