Question

chứng tỏ a/n(n+a) = 1/n - 1/n+a (n,a ϵ số tự nhiên khác 0 ). Áp dụng tính:

A=1/2x3 + 1/3x4 +...+1/99+100

B=5/1x4 + 5/4x7 +...+5/100x103

 

Answer 1

\(A=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)

\(B=\dfrac{5}{1\cdot4}+\dfrac{5}{4\cdot7}+...+\dfrac{5}{100\cdot103}\)

\(=\dfrac{5}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{100\cdot103}\right)\)

\(=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)=\dfrac{5}{3}\cdot\dfrac{102}{103}=\dfrac{170}{103}\)