\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+....+\frac{2}{99\cdot101}\)

\(\frac{2}{1\cdot3}=\frac{3-1}{1\cdot3}=\frac{3}{1\cdot3}-\frac{1}{1\cdot3}=\frac{1}{1}-\frac{1}{3}=1-\frac{1}{3}\)

\(\frac{2}{3\cdot5}=\frac{5-3}{3\cdot5}=\frac{5}{3\cdot5}-\frac{3}{3\cdot5}=\frac{1}{3}-\frac{1}{5}\)

....

\(\frac{2}{99\cdot101}=\frac{101-99}{99\cdot101}=\frac{101}{99\cdot101}-\frac{99}{99\cdot101}=\frac{1}{99}-\frac{1}{101}\)

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)

\(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+\frac{5}{5\cdot7}+...+\frac{5}{99\cdot101}\)

=\(\frac{5}{2}\cdot\frac{2}{1\cdot3}+\frac{5}{2}\cdot\frac{2}{3\cdot5}+\frac{5}{2}\cdot\frac{2}{5\cdot7}+...+\frac{5}{2}\cdot\frac{2}{99\cdot101}\)

=\(\frac{5}{2}\cdot\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right]\)

=\(\frac{5}{2}\cdot\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right]\)

=\(\frac{5}{2}\cdot\left(1-\frac{1}{101}\right)\)

=\(\frac{5}{2}\cdot\frac{100}{101}\)

\(=\frac{250}{101}\)

= 3 - 1 / 1 x 3 + 5 - 3 / 3 x 5 + ... + 101 - 99 / 99 x 101

= 1 - 1 / 3 + 1 / 3 - 1 / 5 + 1 / 5 - ... - 1 / 99 + 1 / 99 - 1 / 101

gạch gạch gạch gạch ... gạch gạch

= 1 - 1 / 101

= 100 / 101

\(\frac{2}{1.2}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\)

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Bạn giúp mk nốt b được ko?

\(A=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)

\(A=5.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{99.101}\right)\)

\(A=5.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{99.101}\right)\)

\(A=\frac{5}{2}.\left(1-\frac{1}{3}+1-\frac{3}{5}+1-\frac{5}{7}+1-\frac{99}{101}\right)\)

\(A=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)

\(A=\frac{5}{2}.\frac{100}{101}=\frac{5.50}{101}=\frac{250}{101}2\frac{48}{101}\)

#Hokrot#

@. C.Ronaldo@ 

Em Sai từ dòng thứ 4 rồi nhé!

A=\(\frac{5}{2}\left(\frac{3-1}{3.1}+\frac{5-3}{3.5}+\frac{7-5}{7.5}+...+\frac{101-99}{101.99}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{101}\right)=\frac{250}{101}\)

\(A=\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{99.101}\)

\(A=5.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\right)\)

\(A=5.\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\right)\)

\(A=\dfrac{5}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(A=\dfrac{5}{2}.\left(1-\dfrac{1}{101}\right)\)

\(A=\dfrac{5}{2}.\dfrac{100}{101}=\dfrac{5.50}{101}=\dfrac{250}{101}=2\dfrac{48}{101}\)

\(\dfrac{250}{101}\)

A = \(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{99.101}\)

= \(\dfrac{5}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\right)\)

= \(\dfrac{5}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

= \(\dfrac{5}{2}.\left(1-\dfrac{1}{101}\right)\)

= \(\dfrac{5}{2}.\dfrac{100}{101}\)

= \(\dfrac{250}{101}\)

Đặt BT trên là A

\(\frac{2}{5}.A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.100}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\)

\(\frac{2}{5}.A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)

\(A=\frac{100}{101}.\frac{5}{2}=\frac{250}{101}\)

Phải là 100/101 : 2/5 mới đúng chứ

A = 1.(1 + 2) + 3.(3 + 2) + 5.(5 + 2) + … + 97.(97 + 2) + 99.(99 + 2)

A = (1² + 3² + 5² + … + 97² + 99²) + 2.(1 + 3 + 5 + … + 97 + 99).

Đặt B = 1² + 3² + 5² + … + 99²

=> B = (1² + 2² + 3² + 4² + … + 100²) – 2².(1² + 2² + 3² + 4² + … + 50²)

Tính dãy tổng quát C = 1² + 2² + 3² + … + n²

C = 1.(0 + 1) + 2.(1 + 1) + 3.(2 + 1) + … + n.[(n – 1) + 1]

C = [1.2 + 2.3 + … + (n – 1).n] + (1 + 2 + 3 + … + n)

C = = n.(n + 1).[(n – 1) : 3 + 1 : 2] = n.(n + 1).(2n + 1) : 6

Áp dụng vào B ta được:

B = 100.101.201 : 6 – 4.50.51.101 : 6 = 166650

=> A = 166650 + 2.(1 + 99).50 : 2

=> A = 166650 + 5000 = 172650.

Vậy: A = 172650.

Chúc bạn học tốt !!!

cái này bạn mở sách bồi dưỡng toán ra trang gần cuối là thấy ngay ấy mà

\(\frac{2}{5}.A\)= \(\frac{2}{1.3}\)+ \(\frac{2}{3.5}\)+ \(\frac{2}{5.7}\)+ ... + \(\frac{2}{99.100}\)= \(\frac{3-1}{1.3}\)+\(\frac{5-3}{3.5}\)+\(\frac{7-5}{5.7}\)+ ... + \(\frac{101-99}{99.101}\)

\(\frac{2}{5}.A\)= 1 \(-\)\(\frac{1}{3}\)+ \(\frac{1}{3}\)\(-\)\(\frac{1}{5}\)+\(\frac{1}{5}\)\(-\)\(\frac{1}{7}\)+ ... + \(\frac{1}{99}\)+\(\frac{1}{101}\)= 1\(-\)\(\frac{1}{101}\)=\(\frac{100}{101}\)

\(A\)=\(\frac{100}{101}\): \(\frac{2}{5}\)= \(\frac{100}{101}\).\(\frac{5}{2}\)= \(\frac{250}{101}\)

=5/2.(2/1.3+2/3.5+2/5.7+...+2/99.101)

=5/2.(1-1/3+1/3-1/5+1/5-1/7+..+1/99-1/101)

=5/2.(1-1/101)

=5/2.100/101

=250/101

\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)

\(=5\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(\Leftrightarrow\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)

\(=\frac{5}{2}.\frac{100}{101}\)

\(=\frac{250}{101}\)