\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{1}{k}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Leftrightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2\)

k=2

chuan 100%

k=2 do

a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)

\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=5\left(1-\dfrac{1}{100}\right)\)

\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)

b, \(C=1.2.3+2.3.4+...+8.9.10\)

\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)

\(C=\dfrac{8.9.10.11}{4}=1980.\)

c, https://hoc24.vn/hoi-dap/question/384591.html

Câu này bạn vào đây mình đã giải câu tương tự nhé.

\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)

\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)

\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)

\(\Leftrightarrow A=\dfrac{99}{20}\)

Gọi \(A=1×2+2×3+..+99×100\)

\(3A=1.2.3+2.3.3+...+999.100.3=1.2\left(3-0\right)+2.3\left(4-1\right)+...+98.99\left(100-97\right)=1.2.3+2.3.4-1.2.3+...-98.99.100-99.100.101=99.100.101\)

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

em chịu

thôi đừng giúp nữa , mình nghĩ RA RỒI

1/1.2 + 1/2.3 + 1/3.4 +......+1/99.100

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

= [ ( -1/2 +1/2) +( -1/3+1/3) + (-1/4 + 1/4) +..... +( -1/99+1/99 ) ] + ( 1/1 + -1/100 ) 

= 0 + 99/100

= 99/100

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

\(=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)

\(=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)

ủng hộ nha

xin đấy k mk 1 cái nhé

\(B=1.2+2.3+....+99.100\)

\(\Rightarrow3B=1.2.3+2.3.4+...+99.100.3\)

\(\Rightarrow3B=1.2.\left(3-0\right)+2.3.\left(4-1\right)+....+99.100.\left(101-98\right)\)

            \(=\left(1.2.3+2.3.4+....+99.100.101\right)-\left(0.1.2+1.2.3+...+98.99.100\right)\)

              \(=99.100.101-0.1.2\)

               = 999900 - 0

=> B = 999900 : 3 = 333300

Vậy B = 333300

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) + ...+ 99.100.(101-98)

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

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

3B = 99.100.101

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

\(B=1.2+2.3+3.4+...+99.100\)

\(\Rightarrow3B=1.2.3+2.3.3+....+99.100.3\)

\(\Rightarrow3B=1.2.3+2.3.(4-1)+...+99.100.(101-98)\)

\(\Rightarrow3B=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100\)

\(\Rightarrow3B=99.100.101\)

\(\Rightarrow B=333300\)

A=1/100

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

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

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

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

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

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

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

\(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100.101}\)

\(S=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{99.100}-\dfrac{1}{100.101}\right)\)

\(S=\dfrac{1}{4}-\dfrac{1}{2.100.101}\)