Question

I. Tính các tổng sau:

a, C = 7/10.11 + 7/ 11.12 +7/12.13 + ... + 7/ 69.70

b, E = 3^2/ 8.11 + 3^2/ 11.14 + 3^2/ 14.17 + .. + 3^2/ 197.200

Cố gắng giúp mjk nhá!!! Thks nhìu lắm <3

Who help me?

Answer 1

a, Ta có :

\(C=\dfrac{7}{10.11}+\dfrac{7}{11.12}+.......+\dfrac{7}{69.70}\)

\(=7\left(\dfrac{1}{10.11}+\dfrac{1}{11.12}+.........+\dfrac{1}{69.70}\right)\)

= \(7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+........+\dfrac{1}{69}-\dfrac{1}{70}\right)\)

= \(7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=7.\dfrac{6}{70}=\dfrac{3}{5}\)

b, \(E=\dfrac{3^2}{8.11}+\dfrac{3^2}{11.14}+........+\dfrac{3^2}{197.200}\)

\(=3.\left(\dfrac{3}{8.11}+\dfrac{3}{11.14}+......+\dfrac{3}{197.200}\right)\)

= \(3.\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+......+\dfrac{1}{197}-\dfrac{1}{200}\right)\)

= \(3.\left(\dfrac{1}{8}-\dfrac{1}{200}\right)=3.\dfrac{24}{200}=\dfrac{9}{25}\)