Question

So sánh 1/1*2+1/2*3+1/3*4+1/4*5+1/5*6+...+1/13*14 và 1

Answer 1

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{13.14}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{13}-\dfrac{1}{14}\)

\(=\dfrac{1}{1}-\dfrac{1}{14}=\dfrac{13}{14}\)

Ta thấy: \(\dfrac{13}{14}< 1\)

Vậy A<1

Answer 2

Đặt \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{13\cdot14}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{13}-\dfrac{1}{14}\)

\(=1-\dfrac{1}{14}\)

\(=\dfrac{14}{14}-\dfrac{1}{14}\)

\(=\dfrac{13}{14}< 1\)

Vậy A < 1

Answer 3

Ta có :B= \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{13.14}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+....+\dfrac{1}{13}-\dfrac{1}{14}\)

\(=1-\dfrac{1}{14}\)

Vì \(1-\dfrac{1}{14}< 1\)

Suy ra : \(B< 1\)

Answer 4

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{13\cdot14}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{13}-\dfrac{1}{14}\)

\(=1-\dfrac{1}{14}=\dfrac{13}{14}< 1\)

\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+....+\dfrac{1}{13\cdot14}< 1\)

Vậy \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{13\cdot14}< 1\)