Ta có : abba = 1001a + 110b 

Mà 1001 chai hết cho 11 và 110 chai hết cho 11

Nên 1001a chia hết cho 11 và 110b chia hết cho11

Suy ra abba chia hết cho 11

Ta có: S = 1.2 + 2.3 + 3.4 + ....... + 99.100 + 100.101

=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ....... + 100.101.102

=> 3S = 100.101.102

=> S = 100.101.102 / 3

=> S = 343400

\(\frac{3}{1.2}+\frac{3}{2.3}+........+\frac{3}{99.100}\)

\(=3\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\right)\)

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

\(=3\left(1-\frac{1}{100}\right)\)

\(=\frac{3.99}{100}=\frac{297}{100}\)

Ta có :

\(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+..............+\dfrac{1}{99.100}\)

\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...........+\dfrac{1}{99}-\dfrac{1}{100}\)

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

\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99x100}\)

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

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

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

Vãi cả nhân :V

\(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+\frac{2}{5\cdot6}\\ =2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\right)\\ =2\left(\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+\frac{5-4}{4\cdot5}+\frac{6-5}{5\cdot6}\right)\\ =2\left(\frac{2}{1\cdot2}-\frac{1}{1\cdot2}+\frac{3}{2\cdot3}-\frac{2}{2\cdot3}+\frac{4}{3\cdot4}-\frac{3}{3\cdot4}+\frac{5}{4\cdot5}-\frac{4}{4\cdot5}+\frac{6}{5\cdot6}-\frac{5}{5\cdot6}\right)\\ =2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...-\frac{1}{6}\right)\\ =2\left(1-\frac{1}{6}\right)\\ =2\cdot\frac{5}{6}=\frac{10}{6}\)

Chúc bạn học tốt nha.

\(B=1+\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}.\)

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

\(B=1+1-\frac{1}{100}=2-\frac{1}{100}\)

\(B=\frac{199}{100}\)

\(C=\frac{1}{1.2}+\frac{1}{2.3}+........+\frac{1}{n\left(n+1\right)}\)

\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.......+\frac{1}{n}-\frac{1}{n+1}\)

\(C=1-\frac{1}{n+1}\)

\(C=\frac{n+1-1}{n+1}=\frac{n}{n+1}\)

Áp dụng công thức tình dãy số ta có :

\(D=\frac{\left[\left(n-1\right):1+1\right].\left(n+1\right)}{2}=\frac{n.\left(n+1\right)}{2}\)

H = \(\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+\frac{1}{3.4}-\frac{1}{3.4.5}+...+\frac{1}{99.100}-\frac{1}{99.100.101}\)

   \(=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{99.100.101}\right)\)

Đặt G = \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)

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

          = \(1-\frac{1}{100}\)

           = \(\frac{99}{100}\)

Đặt K = \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{99.100.101}\right)\)

=>2K = \(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{99.100.101}\right)\)

          = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\)

          = \(\frac{1}{1.2}-\frac{1}{100.101}\)

          = \(\frac{1}{2}-\frac{1}{10100}\)

          = \(\frac{5049}{10100}\)

=> K =\(\frac{5049}{10100}:2=\frac{5049}{10100}.\frac{1}{2}=\frac{5049}{20200}\)

Thay G,K vào H ta có :

H = \(\frac{99}{100}-\frac{5049}{20200}\)

Tự tính :)

\(H=\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+...+\frac{1}{99.100}-\frac{1}{99.100.101}\)

\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.34}+...+\frac{1}{99.100.101}\right)\)

\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{99.100.101}\right)\)

\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)

\(=\frac{99}{100}-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)=\frac{99}{100}-\frac{1}{2}.\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)

sao nhiều vậy!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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

= 1 - 1/100

= 99/100