A = 1.2. + 2.3 + 3.4 + ... + 99.100

3A = 1.2.3 + 2.3.(4-1) + ... + 99.100.(101-98)

3A = 1.2.3 + 2.3.4 - 2.3.1 + ... + 99.100.101 - 99.100.98

3A = 99.100.101

3A = 999900

A = 333300

lấy nick khác hả không qua được mắt tui đâu đồ bất công

A=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

A=1-1/100                            A=99/100                                                                                    B= (1/5.6+1/6/7+...+1/101.102).3                         B=(1/5-1/6+1/6-1/7+...+1/101-1/102).3        B=(1/5-1/102).3                                                 B=97/170                                                            

1) Tính

a) Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)

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

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

lấy 1 chia cho các tổng rồi áp dụng công thức là ra

a) A = 1.3 +2.4 + 3.5 +...+ 97.99 + 98.100

A = 1(2 + 1) + 2(3+1) + 3(4 + 1) +...+ 98(99+1)

= (1.2 + 2.3 + 3.4 +...+ 98.99) + (1 + 2 + 3 +...+ 98)

= [ 1.2.3 + 2.3.(4-1) +...+ 98.99.(100-97)] + [ 1.2 + 2.(3-1) + 3.(4-2) +... 98.(99-97)]

= [ 1.2.3 + 2.3.(4-1) - 1.2.3 + 3.4.(5-2) - 2.3.(4-1) +...+ 98.99.(100-97) - 97.98(99-96)] + [ 1.2 + 2.(3-1) - 1.2 + 3.(4-2) - 2.(3-1) +...+ 98.(99-97) - 97(98-96)]

= 98.99.100:3 + 98.99:2 = 323 400 + 4581 = 328251

b) B = 1.2.3 + 2.3.4 + 3.4.5 +...+ 48.49.50

4B = 1.2.3.4 + 2.3.4.(5-1) + 3.4.5.(6-2) +...+ 48.49.50.(51-47)

4B-B = 1.2.3.4 + 2.3.4.(5-1) - 1.2.3.4 + 3.4.5.(6-2) - 2.3.4.(5-1) +...+ 48.49.50.(51-47) - 47.48.49.(50-46)

= 48.49.50.51:4 = 1499400

c) C = 1/2 + 1/2^2 + 1/2^3 +...+ 1/2^10

=> 2C = 1 + 1/2 + 1/2^2 +...+ 1/2^9 (1)

= 1/2 + 1/2^2 +...+ 1/2^10

Lấy (1)-(2) ta được:

A = 1-1/2^10 = 1024/1024 - 1/1024 = 1023/1024

Chúc bạn học tốt

\(\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}-2x=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-2x=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)\(5\left(1-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(5.\frac{99}{100}-2x=\frac{1}{2}.\frac{98}{99}\)

\(\frac{99}{20}-2x=\frac{49}{99}\)

\(2x=\frac{99}{20}-\frac{49}{99}\)

\(2x=\frac{8821}{1980}\)

\(x=\frac{8821}{1980}:2\)

\(x=\frac{8821}{3960}\)

a)1/5.6+1/6.7+1/7.8+.......+1/99.100

= (1/5-1/6)+(1/6-1/7)+(1/7-1/8)+.....+(1/99-1/100)

= 1/5 - 1/100

= 19/100

b)2/1.3+2/3.5+2/5.7+.........+2/2013.2015

= (1/1-1/3)+(1/3-1/5)+(1/5-1/7)+.....+(1/2013+1/2015)

= 1/1 - 1/2015

= 2014/2015

\(a,\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{99.100}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{5}-\frac{1}{100}=\frac{20}{100}-\frac{1}{100}=\frac{19}{100}\)

\(b,\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\)

\(=\frac{1}{1}-\frac{1}{2015}=\frac{2015}{2015}-\frac{1}{2015}=\frac{2014}{2015}\)

Bài 1 : Ta có : a = 1.2 + 2.3 + 3.4 + ....... + 99.100

=> 3a = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) + ...... + 99.100.(101 - 98)

=> 3a = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 99.100.101

=> 3a = 99.100.101

=>   a = \(\frac{99.100.101}{3}=333300\)