Ta đặt

  \(A=\dfrac{7}{1\times2}+\dfrac{7}{2\times3}+...+\dfrac{7}{99\times100}\)

\(\dfrac{1}{7}\times A=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+....+\dfrac{1}{99\times100}\)

\(\dfrac{1}{7}\times A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\dfrac{1}{7}\times A=1-\dfrac{1}{100}\)

\(\dfrac{1}{7}\times A=\dfrac{99}{100}\)

\(A=\dfrac{99}{100}\div\dfrac{1}{7}\)

\(A=\dfrac{693}{100}\)

= 7.(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100)

= 7.(1 - 1/100)

= 7 . 99/100

= 693/100

\(A=7\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)

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

\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{100-99}{99.100}=\)

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

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

\(\Rightarrow A=7x\dfrac{99}{100}=6,93\)

\(1.2^2+2.3^2+...+99.100^2\)

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

\(=1.2.3-1.2+2.3.4-2.3+...+99.100.101-99.100\)

\(=\left(1.2.3+2.3.4+...+99.100.101\right)\)\(-\left(1.2+2.3+...+99.100\right)\)

Chúc học tốt

Ta có : \(S=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{99.100}\right)\)

\(\Rightarrow S=2.\left(1-\frac{1}{100}\right)\)

\(=2.\frac{99}{100}=\frac{99}{50}\)

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

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

S= 2.\(\frac{99}{100}\)

S=\(\frac{99}{50}\)

\(S=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{98.99}+\frac{2}{99.100}\)

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

\(S=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(S=2.\left(1-\frac{1}{100}\right)\)

\(S=2.\frac{99}{100}\)

\(S=\frac{99}{50}\)

Ta thấy mỗi tổng trên là tích của hai số tự nhiên liên tiếp. 
\(a_1=1.2\Rightarrow3a_1=1.2.3\)\(\Rightarrow3a_1=1.2.3-0.1.2\).
\(a_2=2.3\Rightarrow3a_2=2.3.3\)\(\Rightarrow3a_2=2.3.4-1.2.3\).
.....
\(a_{99}=99.100\Rightarrow3a_{99}=3.99.100\)\(\Rightarrow3a_{99}=98.99.100-97.98.99\).
Ta có:
\(3A=1.2.3+2.3.3+3.4.3+....+99.100.3\)
\(=\)\(1.2.3-0.1.2+2.3.4-1.2.3+........+98.99.100-97.98.100\)
\(=98.99.100\)
Suy ra: \(A=\frac{98.99.100}{3}=323400\).

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

⇒3B=1.2.3+2.3.3+....+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=99.100.101

\(⇒\)

A = 1 . 2 + 2 . 3 + 3 . 4 + ... + 99 . 100

3A = 1 . 2 . 3 + 2 . 3 . 3 + 3 . 4 . 3 + ... + 99 . 100 . 3

3A = 1 . 2 . 3 + 2 . 3 . ( 4 - 1 ) + 3 . 4 . ( 5 - 2 ) + ... + 99 . 100 . ( 101 - 98 )

3A = 1 . 2 . 3 + 2 . 3 . 4 - 1 . 2 . 3 + 3 . 4 . 5 - 2 . 3 . 4 + ... + 99 . 100 . 101 - 98 . 99 . 100

3A = 99 . 100 . 101

A = 99 . 100 . 101 : 3 = 333300

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

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

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

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

              \(=99.100.101-0.1.2\)

               = 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

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

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

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

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

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

\(\Rightarrow3B=99.100.101\)

\(\Rightarrow B=333300\)

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

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

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

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

nhanh lên nha mấy bn ai trả lời dcd thì kb với mình nha

thank you

\(B=1\cdot2^2+2\cdot3^2+3\cdot4^2+...+99\cdot100^2\\ =\left(2-1\right)\cdot2^2+\left(3-1\right)\cdot3^2+\left(4-1\right)\cdot4^2+...+\left(100-1\right)\cdot100^2\\ =2\cdot2^2-1\cdot2^2+3\cdot3^2-1\cdot3^2+4\cdot4^2-1\cdot4^2+...+100\cdot100^2-1\cdot100^2\\ =2^3-2^2+3^3-3^2+4^3-4^2+...+100^3-100^2\\ =1^3-1^2+2^3-2^2+3^3-3^2+4^3-4^2+...+100^3-100^2\\=\left(1^3+2^3+3^3+4^3+...+100^3\right)-\left(1^2+2^2+3^2+4^2+...+100^2\right)\\=\left(1+2+3+...+100\right)^2-\dfrac{100\cdot\left(100+1\right)\cdot\left(2\cdot100+1\right)}{6}\\ =\left[\dfrac{\left(100\cdot101\right)}{2}\right]^2-\dfrac{100\cdot101\cdot201}{6}\\ =5050^2-2030100\\ =25502500-2030100\\ =23472400 2\)

Bonus: Công thức:

1) \(1+2+3+...+n=\dfrac{n\cdot\left(n+1\right)}{2}\)

2) \(1^2+2^2+3^2+...+n^2=\dfrac{n\cdot\left(n+1\right)\cdot\left(2n+1\right)}{6}\)

3) \(1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\)

kết quả dư số 2 nhé bạn, bỏ đi

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

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

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

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

k=2

chuan 100%

k=2 do