Đặt tổng trên = A

Có : 3A = 1.2.3+2.3.3+....+98.99.3

 = 1.2.3+2.3.(4-1)+.....+98.99.(100-97)

 = 1.2.3+2.3.4-1.2.3+.....+98.99.100-97.98.99

 = 98.99.100

=> A = 98.99.100/3 = 323400

k mk nha

Gọi A = 1.2 + 2.3 + .. + 98.99

3A = 1.2.3 + 2.3.3 + ... + 98.99.3

3A = 1.2.3 + 2.3.(4 - 1) + ... + 98.99.(100 - 97)

3A = 1.2.3 + 2.3.4 - 1.2.3 + ... + 98.99.100 - 97.98.99

3A = 98.99.100

3A = 970200

A = 323400

\(A=1.2^2+2.3^2+...+98.99^2\)

\(=1.2.\left(3-1\right)+2.3.\left(4-1\right)+...+98.99.\left(100-1\right)\)

\(=1.2.3-1.2+2.3.4-2.3+...+98.99.100-98.99\)

\(=\left(1.2.3+2.3.4+...+98.99.100\right)-\left(1.2+2.3+...+98.99\right)\)

\(=\dfrac{98.99.100.101}{4}+\dfrac{98.99.100}{3}\)

\(=24497550+323400\)

\(=24820950\)

Đặt \(A=1+\left(1+2\right)+\left(1+2+3\right)+\cdots+\left(1+2+3+\cdots+98\right)\)

\(=\frac{1\cdot2}{2}+\frac{2\cdot3}{2}+\frac{3\cdot4}{2}+\cdots+\frac{98\cdot99}{2}\)

\(=\frac{1\cdot2+2\cdot3+\cdots+98\cdot99}{2}\)

Ta có: \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+\cdots+\left(1+2+3+\cdots+98\right)}{1\cdot2+2\cdot3+\cdots+98\cdot99}\)

\(=\frac{\frac{1\cdot2+2\cdot3+\cdots+98\cdot99}{2}}{1\cdot2+2\cdot3+\cdots+98\cdot99}=\frac12\)

A=98.99.100/3 = 323400

B=98.99/2       =4851