B = 1.2+2.3+3.4+...+99.100

B=1.100

B=100

C=1.3+2.4+3.5+4.6+...+9.11

C=1.(2+1)+2.(3+1)+3.(4+1)+4.(5+1)+...+9.(10+1)

C=1.2+1+2.3+1+3.4+1+4.5+1+...+9.10+1

C=(1.2+2.3+3.3+4.5+...+9.10)+(1+1+1+1+..+1)

C=1.10+10

C=10+10

C=20

a) B = 1.2+2.3+3.4+..+99.100

=>3B=1.2.3+2.3.3+3.4.3+...+99.100.3

3B = 1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)

3B = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5-2.3.4+...+99.100.101-98.99.100

3B = (1.2.3+2.3.4+3.4.5+..+99.100.101) - (1.2.3+2.3.4+...+98.99.100)

3B = 99.100.101

\(B=\frac{99.100.101}{3}=333300\)

b) C = 1.3+2.4+3.5+4.6+...+9.11

C = (2-1).(2+1)+(3-1).(3+1) + (4-1).(4+1)+(5-1).(5+1)+...+(10-1).(10+1)

C = 2² - 1 + 3² - 1 + 4² - 1 + 5² - 1 +...+10² - 1

C = (2²+3²+4²+5²+...+10²) -(1+1+...+1) 

...

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

    =1-1/101

    =100/101

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

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

    =(1-1/101).2,5

    =100/101.2,5

    =250/101

c) =(2/2.4+2/4.6+2/6.8+...+2/2008-2/2010).2

    =(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010).2

    =(1/2-1/2010).2

    =1004/1005

Ta có  \(A=\dfrac{2}{1.3}-\dfrac{2}{2.4}+\dfrac{2}{3.5}-\dfrac{2}{4.6}+\dfrac{2}{5.7}-\dfrac{2}{6.8}+\dfrac{2}{7.9}-\dfrac{2}{8.10}+\dfrac{2}{9.11}-\dfrac{2}{10.12}\) 

\(\Rightarrow A=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\right)-\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+\dfrac{2}{8.10}+\dfrac{2}{10.12}\right)\) \(\Rightarrow A=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)-\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}\right)\) \(\Rightarrow A=\left(1-\dfrac{1}{11}\right)-\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\) 

\(\Rightarrow A=1-\dfrac{1}{11}-\dfrac{1}{2}+\dfrac{1}{12}\) 

\(\Rightarrow A=\dfrac{9}{22}+\dfrac{1}{12}\) 

\(\Rightarrow A=\dfrac{65}{132}\) 

Mà \(\dfrac{65}{132}< 1\) \(\Rightarrow A< 1\) 

Vậy \(A< 1\)

\(A=\frac{2^2}{1.3}\cdot\frac{2^2}{2.4}\cdot\frac{2^2}{3.5}\cdot\frac{2^2}{4.6}\)

\(A=\frac{4}{3}\cdot\frac{1}{2}\cdot\frac{4}{15}\cdot\frac{1}{6}\)

\(A=\frac{4.1.4.1}{3.2.15.6}\)

\(A=\frac{4}{135}\)

\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)

\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}\)

\(=\frac{2.3.4.5}{1.2.3.4}.\frac{2.3.4.5}{3.4.5.6}\)

\(=\frac{5}{1}.\frac{2}{6}\)

\(=\frac{5}{1}.\frac{1}{3}\)

\(=\frac{5}{3}\)

Đọc kỹ lại đề đi Trần Cao Vỹ Lượng

C=2.2.3.3.4.4.5.5/1.3.2.4.3.5.4.6

C=(2.3.4.5).(2.3.4.5)/(1.2.3.4).(3.4.5.6)

C=2.5/6

C=5/3

C=2^2/1.3.3^2/2.4.5^2/3.5.4^2/4.6

C= 2.2.3.3.5.5.4.4/1.3.2.4.3.5.4.6

C=(2.3.4.5).(2.3.4.5)/(1.2.3.4).(3.4.5.6)

C=5.2/6

C=5/3

\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{5^2}{3.5}.\frac{4^2}{4.6}\)

\(=\frac{2^2.3^2.5^2.4^2}{1.3.2.4.3.5.4.6}\)

\(=\frac{2^2.3^2.4^2.5^2}{1.2.3^2.4^2.5.6}\)

\(=\frac{2.5}{6}\)

\(=\frac{10}{6}=\frac{5}{2}\)

\(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}......\frac{10^2}{9.11}=\frac{\left(1.2.3.....10\right)^2}{\left(1.2.3.....9\right).\left(3.4.5....9.10.11\right)}=\frac{\left(1.2.3....10\right)^2}{\left(1.2\right)\left(3.4.5.....9\right)^2\left(10.11\right)}=\frac{\left(1.2.10\right)^2}{\left(1.2\right).\left(10.11\right)}=\frac{1.2.10}{11}=\frac{20}{11}\)

\(\dfrac{4}{3}\times\dfrac{9}{8}\times\dfrac{16}{15}\times\dfrac{25}{24}=\dfrac{5}{3}\)

`(2^2)/(1 . 3) . (3^2)/(2 . 4) . (4^2)/(3 . 5) . (5^2)/(4 . 6)`

`= 4/3 . 9/8 . 16/15 . 25/24 = 5/3`

5/3