Question

[Uzumaki Naruto]

E=2/1x4  +   2/4x7  +   2/7x10    +......+2/100x103

Giải đầy đủ

Answer 1

[Arima Kousei]

Dấu \(.\)là dấu nhân 

Ta có : 

\(E=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{100.103}\)

\(\Rightarrow E=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{2}{100.103}\right)\)

\(\Rightarrow E=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(\Rightarrow E=\frac{2}{3}.\left(1-\frac{1}{103}\right)\)

\(\Rightarrow E=\frac{2}{3}.\frac{102}{103}\)

\(\Rightarrow E=\frac{68}{103}\)

Vậy \(E=\frac{68}{103}\)

~ Ủng hộ nhé 

Answer 2

[Trần Thanh Phương]

\(E=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+...+\frac{2}{100\cdot103}\)

\(E=2\cdot\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{100\cdot103}\right)\)

Gọi tổng trong ngoặc là F

\(\Rightarrow3F=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{100\cdot103}\)

\(\Rightarrow3F=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)

\(\Rightarrow3F=1-\frac{1}{103}=\frac{102}{103}\)

\(\Rightarrow F=\frac{102}{103\cdot3}=\frac{34}{103}\)

\(\Leftrightarrow E=2\cdot\frac{34}{103}=\frac{68}{103}\)

Vậy......