Question

1.Tính tổng các số hạng trên:

\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{100.102}\)

2.Cho P=\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{33}\)

Chứng minh\(\frac{3}{5}< P< \frac{4}{5}\) 

Answer 1

1. 

\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{100.102}\)

\(=\frac{4-2}{2.4}+\frac{6-4}{4.6}+....+\frac{102-100}{100.102}\)

\(=\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{100}-\frac{1}{102}\right)\times\frac{1}{2}\)

\(=\left(\frac{1}{2}-\frac{1}{102}\right)\times\frac{1}{2}\)

\(=\frac{25}{51}\times\frac{1}{2}\)

\(=\frac{25}{102}\)

Answer 2

1,

\(A=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{100.102}\)

\(2A=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{100.102}\)

\(2A=\frac{4-2}{2.4}+\frac{6-4}{4.6}+...+\frac{102-100}{100.102}\)

\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{102}\)

\(2A=\frac{1}{2}-\frac{1}{102}\)

\(2A=\frac{25}{51}\)

\(A=\frac{25}{102}\)

2,câu hỏi tương tự

Answer 3

\(a=\frac{1}{2.4}+\frac{1}{4.6}+....+\frac{1}{100.102}\)

\(2a=\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{1}{100.102}\)

Answer 4

đặt a=1/2x4+1/4x6+...+1/100x102

a x2=(1/2x4+1/4x6+...+1/100x102)x2

a x2=2/2x4+2/4x6+...+2/100x102

a x2=1/2-1/4+1/4-1/6+...+1/100-1/102

khử đi ta còn

a x2=1/2-1/102

a x2=25/51

a=25/51 :2=25/102