hic hic mai phải nộp rồi mọi người giúp kim oanh nha oaoa

=1.99+2.(99-1)+3.(99-2)+...+98.(99-97)+99(99-98)

=99.(1+2+3+4+...+98+99)-(2+2.3+3.4+...+97.98+98.99)

=99.(1+99).99/2-98.99.100/3

=99.50.99-98.33.100

=490050-323400

=166650

\(C=1.99+2.98+3.97+........+97.3+98.2+99.1\)

\(\Rightarrow C=1.99+2.\left(99-1\right)+3\left(99-2\right)+..........+98.\left(99-97\right)+99.\left(99-98\right)\)

\(\Rightarrow C=1.99+2.99-1.2+3.99-2.3+........+98.99-97.98+99.99-98.99\)

\(\Rightarrow C=\left(1.99+2.99+.......+99.99\right)-\left(1.2+2.3+.........+98.99\right)\)

\(\Rightarrow C=490050-\left(1.2+2.3+....+98.99\right)\)

Đặt \(A=1.2+2.3+3.4+........+98.99\)

\(\Rightarrow3A=1.2.3+2.3.3+..........+98.99.3\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+.....+98.99\left(100-97\right)\)

\(\Rightarrow3A=1.2.3-1.2.3+2.3.4-2.3.4+......+97.97.99-97.98.99+98.99.100\)

\(\Rightarrow3A=98.99.100\)

\(\Rightarrow A=\dfrac{98.99.100}{3}=323400\)

\(\Rightarrow C=490050-323400=166650\)

Vậy \(C=166650\)

Xem lại đề đi bạn. Đề có vấn đề?!

\(A = 1.99 + 2.98 + 3.97 + ...+ 97.3 + 98.2 + 99.1\)

\(A=1.99+2.\left(99-1\right)+3.\left(99-2\right)+...+98.\left(99-97\right)+99.\left(99-98\right)\)

\(A=1.99+2.99-1.2+3.99-2.3+98.99-97.98+99.99-98.99\)

\(=\left(1.99+2.99+3.99+...+98.99+99.99\right)-\left(1.2+2.3+3.4+...+97.98+98.99\right)\)

\(=99.\left(1+2+3+...+98+99\right)-\left(1.2+2.3+3.4+...+97.98+98.99\right)\)

\(=99.4950-\left(1.2+2.3+3.4+97.98+98.99\right)\)

Mà \(1.2+2.3+3.4+...97.98+98.99\)

\(=\frac{1}{3}.\left[1.2+2.3.\left(4-1\right)+3.4.\left(5-2\right)+98.99.\left(100-97\right)\right]\)

\(=\frac{1}{3}.98.99.100=323400\)

\(\Rightarrow A=99.4950-323400=166650\)

Tick cho nè

bo tay

Ô tô đi với vận tốc 50km/giờ vì :

         100 : 2 = 50

                   đs : 50

Ta có: \(1\cdot99^2+2\cdot98^2+\cdots+99\cdot1^2\)

\(=99^2\left(100-99\right)+98^2\left(100-98\right)+\cdots+1^2\left(100-1\right)\)

\(=100\left(1^2+2^2+\cdots+99^2\right)-\left(1^3+2^3+\cdots+99^3\right)\)

\(=100\cdot\frac{99\left(99+1\right)\left(2\cdot99+1\right)}{6}-\left(1+2+\cdots+99\right)^2\)

\(=100\cdot\frac{99\cdot100\cdot199}{6}-\left\lbrack99\cdot\frac{100}{2}\right\rbrack^2\)

\(=100\cdot33\cdot50\cdot199-\left(99\cdot50\right)^2\)

\(=5000\cdot33\cdot199-99^2\cdot50^2=8332500\)

Ô tô đi với vận tốc 50km/giờ vì :

         100 : 2 = 50

                   đs : 50

Ta có: C = 1.2 + 2.3 + 3.4 + ..... + 98.99

=> 3C = 3.(1.2 + 2.3 + 3.4 + ..... + 98.99)

=> 3C = 1.2.(3 - 0) + 2.3.(4 - 1) + .... + 98.99.(100 - 97)

=> 3C = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 98.99.100

=> 3C = 98.99.100

=> C = $C=\frac{98.99.100}{3}=485100$