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

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{100}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)\(=\frac{101}{101}-\frac{1}{101}\)

\(=\frac{100}{101}\)

thank you

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

=\(\frac{1}{1}\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+............+\(\frac{1}{100}\)-\(\frac{1}{101}\)

=1-\(\frac{1}{101}\)=\(\frac{100}{101}\)

\(\sqrt{1.2015}\le\frac{2016}{2}\Rightarrow\frac{1}{\sqrt{1.2015}}\ge\frac{2}{2016}\)

=>S\(\ge\frac{2.1015}{2016}\)\(>\frac{2.2014}{2015}\)

Bài này dễ dùng Cô si cho mẫu là OK

=\(11\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99\cdot100}\right)\)=\(11\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)=\(11\left(1-\frac{1}{100}\right)\)=11\(\frac{99}{100}\)=\(\frac{1089}{100}\)

Đặt \(A=\frac{11}{1.2}+\frac{11}{2.3}+...+\frac{11}{99.100}\)

\(\Rightarrow A=11\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Rightarrow A=11\left(1-\frac{1}{100}\right)\)

\(\Rightarrow A=11.\frac{99}{100}\)

\(\Rightarrow A=\frac{1089}{100}\)

11/1.2+11/2.3+.....+11/99.100

.=11.(1/1.2+1/2.3+....+1/99+100)

=11.(1-1/2+1/2-1/3+.....+1/99-1/100)

=11.(1-1/100)

=11.99/100

=1089/100

(1-1/1.2)+(1-1/2*3)+......+(1-1/2015*2016)

=(0/1*2)+(0+2*3)+..........+(0/2015*2016)

=0

tui nghĩ cái đề phải như thế này  \(\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+\left(1-\frac{1}{3.4}\right)+...+\left(1-\frac{1}{2015.2016}\right)\)

=0 con minh ko giai

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2018\cdot2019}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\)

\(A=1-\frac{1}{2019}=\frac{2018}{2019}\)

Mà \(\frac{2018}{2019}< \frac{2019}{2019}=1\)

\(\Rightarrow A< 1\)

#)Giải :

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)

\(A=1-\frac{1}{2019}\)

\(A=\frac{2018}{2019}\)

Vì \(\frac{2018}{2019}< 1\Rightarrow A< 1\)

      #~Will~be~Pens~#

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2018.2019}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\)

\(=1-\frac{1}{2019}< 1\)

\(\Leftrightarrow A< 1\)

Ta thấy mỗi số hạng của tổng trên là tích của hai số tự nhên liên tiếp, khi đó: 

Gọi a₁ = 1.2 → 3a₁ = 1.2.3 → 3a₁ = 1.2.3 - 0.1.2
      a₂ = 2.3 → 3a₂ = 2.3.3 → 3a₂ = 2.3.4 - 1.2.3
      a₃ = 3.4 → 3a₃ = 3.3.4 → 3a₃ = 3.4.5 - 2.3.4
      …………………..
      a_n-1 = (n - 1)n → 3a_n-1 =3(n - 1)n → 3a_n-1 = (n - 1)n(n + 1) - (n - 2)(n - 1)n
      a_n = n(n + 1) → 3a_n = 3n(n + 1) → 3a_n = n(n + 1)(n + 2) - (n - 1)n(n + 1)

Cộng từng vế của các đẳng thức trên ta có:

3(a₁ + a₂ + … + a_n) = n(n + 1)(n + 2)

A=1.2+2.3+3.4+........+98.99 
3A=1.2.3+2.3.3+3.4.3+........+98.99.3 
3A=1.2.3+2.3.(4 -1) +3.4.(5 -2)+........+98.99.(100 -97) 
3A=1.2.3+2.3.4 -1.2.3 +3.4.5 -2.3.4 +........+98.99.100 -97.98.99 
3A=98.99.100 
===>A=(98.99.100)/3

#Japhkiel#

A=1.2+2.3+3.4+........+98.99 
3A=1.2.3+2.3.3+3.4.3+........+98.99.3 
3A=1.2.3+2.3.(4 -1) +3.4.(5 -2)+........+98.99.(100 -97) 
3A=1.2.3+2.3.4 -1.2.3 +3.4.5 -2.3.4 +........+98.99.100 -97.98.99 
3A=98.99.100 
A=\(\frac{98.99.100}{3}=\frac{970200}{3}=323400\)

Đặt A\(=1\cdot2+2\cdot3+3\cdot4+......+98\cdot99\), ta có

\(3A=1\cdot2\cdot3+2\cdot3\cdot3+.....+98\cdot99\cdot3\)

\(=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+....+98\cdot99\cdot\left(100-97\right)\)

\(=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+....+98\cdot99\cdot100-97\cdot98\cdot99\)

\(=98\cdot99\cdot100\)

\(\Rightarrow A=\frac{98\cdot99\cdot100}{3}=323400\)