Question

Tính bằng cách hợp lý

a) \(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{110}\)

b) \(B=\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{40}+...+\dfrac{1}{1280}\)

Nguyễn Trần Thành Đạt ; Doan Minh Cuong ; Thái Văn Đạt ; Cẩm Vân Nguyễn Thị ; phynit

Answer 1

A = \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{110}\)

= \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{10.11}\)

= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{11}\)

= \(1-\dfrac{1}{11}\)

= \(\dfrac{10}{11}\)

Vậy A = \(\dfrac{10}{11}\)

Answer 2

a) \(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{110}\)

\(\Leftrightarrow A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{10.11}\)

\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{10}-\dfrac{1}{11}\)

\(\Leftrightarrow A=1-\dfrac{1}{11}=\dfrac{10}{11}\)

Answer 3

a) \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{10.11}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{11}\)

\(A=1-\dfrac{1}{11}\)

\(A=\dfrac{10}{11}\)

b) \(B=\dfrac{1}{2^0.5}+\dfrac{1}{2^1.5}+\dfrac{1}{2^2.5}+\dfrac{1}{2^3.5}+...+\dfrac{1}{2^8.5}\)

\(5B=\dfrac{1}{2^0}+\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^8}\)

\(5B=2-1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...+\dfrac{1}{128}+\dfrac{1}{256}\)

\(5B=2-\dfrac{1}{256}\)

\(5B=\dfrac{511}{256}\)

\(B=\dfrac{511}{256}:5\)

\(B=\dfrac{511}{1280}\)