Question

a) 1/3-1/4

b) 1/4-1/5

c) S= 1/1.2+ 1/2.3+.........+1/N.(N+1)

d) M= 2/3.5+ 2/5.7+....+ 2/97.99

giải thích cho mik luôn ạ

 

Answer 1

a) 

\(\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{4}{12}-\dfrac{3}{12}=\dfrac{1}{12}\)

b)\(\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{5}{20}-\dfrac{4}{20}=\dfrac{1}{20}\)

Answer 2

c) Ta có: \(S=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{n\cdot\left(n+1\right)}\)

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

\(=1-\dfrac{1}{n+1}\)

\(=\dfrac{n+1-1}{n+1}=\dfrac{n}{n+1}\)

Answer 3

d)

 \(M=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\\ =\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{99-97}{97.99}=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\\ =\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)