CÁI NÀY NẾU CÓ PHÂN SỐ THÌ LÀM DỄ HƠN NÈ 

S = 1.2 + 2.3 + 3.4 + ... + 97.98

=> 3S = 1.2.3 + 2.3.3 + 3.4.3 + ... + 97.98.3

           = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 97.98.(99 - 96)

           = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 97.98.99 - 96.97.98

           = 97.98.99 

           = 941 094     

=> S = 941 094 : 3  = 313698

Vậy S = 313698

Đặt 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 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100

3A = 99.100.101

A = 33.100.101

A = 333300

\(A=1.2+2.3+3.4+4.5+...+97.98+98.99+99.100\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+4.5.\left(6-3\right)+...+99.100.\left(101-98\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100\)

\(3A=99.100.101\)

\(A=\frac{99.100.101}{3}=\frac{999900}{3}=333300\)

A = 1 - \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}...-\dfrac{1}{97.98}\)

A= 1-\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{97.98}\right)\)

A= 1- \(\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}...+\dfrac{1}{97}-\dfrac{1}{98}\right)\)

A= 1- \(\left(\dfrac{1}{1}-\dfrac{1}{98}\right)\)

A=1-  1 + \(\dfrac{1}{98}\)

A= \(\dfrac{1}{98}\)

Lời giải:

$1-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{97.98}$

$1-A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{98-97}{97.98}$

$1-A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}$

$=1-\frac{1}{98}$

$\Rightarrow A=\frac{1}{98}$

\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}\)

\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{99-98}{98.99}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}\)

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

\(A=\dfrac{2021}{2022}=\dfrac{2022-1}{2022}=1-\dfrac{1}{2022}\)

Có \(2022>99>0\Leftrightarrow\dfrac{1}{99}>\dfrac{1}{2022}\)

Suy ra \(A>B\).

\(\dfrac{-4}{99}\)

-4/99

Đặt A=1.98+2.97+3.96+...+96.3+97.2+98.1

       B=1.2+2,3+3.4+...+96.97+97.98+98.99

Ta có: A=1+(1+2)+...+(1+2+3+...+97+98)

              =\(\dfrac{1.2}{2}+\dfrac{2.3}{2}+...+\dfrac{98.99}{3}\)

              =\(\dfrac{1.2+2.3+3.4+4.5+...+98.99}{2}\)=\(\dfrac{B}{2}\)

    =>E=\(\dfrac{B}{2}\):2=\(\dfrac{1}{2}\)