Nhầm ,chỉ có một + 1/3.5 thôi các bạn nhé

b: 6B=2*4*6+4*6*6+6*8*6+...+46*48*6+48*50*6

=2*4*6-2*4*6+4*6*8-4*6*8+...-44*46*48+46*48*50-46*48*50+48*50*52

=48*50*52

=>B=20800

d: 9D=1*4*9+4*7*9+...+46*49*9

=1*4*2+1*4*7-1*4*7+1*7*10-1*7*10+...+46*49*52-46*49*43

=1*2*4+46*49*52

=117216

=>D=13024

a: 

Ta có: \(A=1\cdot3+2\cdot4+\cdots+99\cdot101\)

\(=1\left(1+2\right)+2\left(2+2\right)+\cdots+99\left(99+2\right)\)

\(=\left(1^2+2^2+\cdots+99^2\right)+2\left(1+2+\cdots+99\right)\)

\(=\frac{99\left(99+1\right)\left(2\cdot99+1\right)}{6}+2\cdot\frac{99\cdot\left(99+1\right)}{2}\)

\(=\frac{99\cdot100\cdot199}{6}+99\cdot100=33\cdot50\cdot199+99\cdot100\)

\(=33\cdot50\cdot\left(199+3\cdot2\right)=33\cdot50\cdot205=338250\)

đúng thật minh ngu quá nhưng mà mik đần hơn

A = 1.3 + 2.4 + 3.5 + ... + 99.101

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

A = 1.2 + 1 + 2.3 + 2 + 3.4 + 3 + ... + 99.100 + 99

A = (1.2 + 2.3 + 3.4 + ... + 99.100) + (1 + 2 + 3 + ... + 99)

A = 333300 + 4950

a = 338250

ban chi can nhan vao day https://olm.vn/hoi-dap/question/184646.html

A=1.3+2.4+3.5+..........+99.101

A=(2-1).(2+1)+(3-1).(3+1)+......+(100-1).(100+1)

A=2^2-1+3^2-1+..........+100^2-1

A=(2^2+3^2+4^2+..........+100^2)-(1+1+........+1)

A=(2^2+3^2+4^2+..........+100^2)-99

 Còn  lại bạn  tự làm nha

thanks Ngô Cao Hoàng nha!

= 1.( 2 + 1 ) + 2 ( 3 + 1 ) + 3( 4 + 1 ) + ..... + 98( 99 + 1 )

 = 1 . 2 + 1 + 2 . 3 + 2 + 3 . 4 + 3 + ...... 98 . 99 + 98

= ( 1 . 2 + 2 . 3 + 3 . 4 + ..... + 98 . 99 ) + ( 1 + 2 + 3 + ..... + 98 )

= 323400 + 4851 

= 328351

bn ủng hộ vũ khánh linh nha

Ta có: \(A=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\cdot\ldots\cdot\left(1+\frac{1}{2017\cdot2019}\right)\)

\(=\left(1+\frac{1}{\left(2-1\right)\left(2+1\right)}\right)\left(1+\frac{1}{\left(3-1\right)\left(3+1\right)}\right)\cdot\ldots\cdot\left(1+\frac{1}{\left(2018-1\right)\left(2018+1\right)}\right)\)

\(=\frac{2^2-1+1}{\left(2-1\right)\left(2+1\right)}\cdot\frac{3^2-1+1}{\left(3-1\right)\left(3+1\right)}\cdot\ldots\cdot\frac{2018^2-1+1}{\left(2018-1\right)\left(2018+1\right)}\)

\(=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\ldots\cdot\frac{2018^2}{2017\cdot2019}=\frac{2\cdot3\cdot\ldots\cdot2018}{1\cdot2\cdot\ldots\cdot2017}\cdot\frac{2\cdot3\cdot\ldots\cdot2018}{3\cdot4\cdot\ldots\cdot2019}\)

\(=\frac{2018}{1}\cdot\frac{2}{2019}=\frac{4036}{2019}<\frac{4038}{2019}\)

=>A<2