1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10

= 1-1/10

=9/10

bn trân ơi, sao ra 1- (1/10) hay zậy?

Ta có: A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)

   =>  3A = 1.2.(3-0) + 2.3.(4-1) + .... + n.(n+1).(n+2 - n+1)

   => 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + n.(n+1).(n+2)

  =>  3A = n.(n+1).(n+2)

  = > A = \(\frac{\text{n.(n+1).(n+2)}}{3}\)

\(A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

Hình như bn viết sai đề,là 1/x.(x+1) chứ

ukm mik xin lỗi mik viết sai đề đó

Đề sai nhé phải là x(x+1)

Ta có\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2015}{2016}\Leftrightarrow\frac{x}{x+1}=\frac{2015}{2016}\Rightarrow x=2015\)

Vậy \(x=2015\)

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=\left(1-\frac{1}{50}\right)+\left(\frac{-1}{2}+\frac{1}{2}+\frac{-1}{3}+\frac{1}{3}+...+\frac{1}{49}\right)\)

\(A=\frac{49}{50}+0\)

\(A=\frac{49}{50}\)

A=1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9

A=1/3-1/9

A=2/9

các câu 2;3 còn lại giống câu 1 bạn nhé

bạn thay số vào rồi làm tương tự

A=1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9

A=1/3-1/9

A=2/9.

Ai giúp đi, làm ơnnnnnnnnnnnnnnnnnnn

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

\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(B< \frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)

\(B< \frac{50}{60}\Leftrightarrow B< \frac{5}{6}\)

= 2-1/1.2 + 3-2/2.3 + 4-3/3.4 + ...... + 3024-3023/3023.3024

= 1-1/2+1/2-1/3+1/3-1/4+.....+1/3023-1/3024

= 1- 1/3024 = 3023/3024

=1-1/2+1/2-1/3+1/3-1/4+.......+1/3023-1/3014

=1-1/3024=3023/3024

k cho mình nha

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{3023.3024}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{3023}-\frac{1}{3024}\)

\(=1-\frac{1}{3024}\)

\(=\frac{3023}{3024}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)

=1-1/20

=19/20

Đặt S= 1.2 + 2.3 + 3.4 + ...+ 99.100
 3S = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3
3S= 1.2.3+2.3(4-1)+3.4(5-2)+...+98.99(100-97)+99.100(101-98)
3S= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-97.98.99+99.100.101-98.99.100
3S = 99.100.101  3S = 3.33.100.101 
 S=33.100.101= 333300

Bạn rút gọn chéo đi 2 với 2 ,3 với 3 cứ như thế còn mỗi 1/100. k nhé

Ta có: C = 1.2 + 2.3 + 3.4 + ... + 99.100

=> 3C=3 (1.2 + 2.3 + 3.4 + ... + 99.100)

=> 3C = 1.2.(3-0)+2.3.(4-1)+....+99.100.(101-98)

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

=> 3C=99.100.101

=> C=99.100.101/3=333300

Nha bạn