Đặt \(A_n=1\cdot2+2\cdot3+3\cdot4+...+n\cdot\left(n+1\right)\)

Như vậy thì \(3A_n=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+n\left(n+1\right)\left[n+2-\left(n-1\right)\right]=n\left(n+1\right)\left(n+2\right)\)

Do đó \(A_n=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

Gọi số phải tính là S, ta có:

\(S=\frac{1\cdot98+2\cdot97+3\cdot96+...+98\cdot1}{1\cdot2+2\cdot3+3\cdot4+...+98\cdot99}\)

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

\(S=\frac{100\cdot\left(1+2+3+...+98\right)-A_{98}}{A_{98}}=\frac{100\cdot99\cdot49}{A_{98}}-1=\frac{100\cdot99\cdot49}{98\cdot99\cdot100:3}-1=\frac{3}{2}-1=\frac{1}{2}\)

Vậy dãy trên có giá trị là \(\frac{1}{2}\)

A =  \(\frac{1x98+2x97+3x96+...+98x1}{1x2+2x3+3x4+...+98x99}\)

A = \(\frac{1x\left(100-2\right)+2x\left(100-3\right)+3x\left(100-4\right)+...+98x\left(100-99\right)}{1x2+2x3+3x4+...+98x99}\)

A =\(\frac{1x100-1x2+2x100-2x3+3x100-3x4+...+98x100-98x99}{1x2+2x3+3x4+...+98x99}\)

A =\(\frac{100x\left(1+2+3+...+98\right)}{1x2+2x3+3x4+...+98x99}\)  - 1

Ta có: 1 + 2 + 3 + ... + 98

        = 98 x 99 : 2

        =    9702  : 2

        =           4851

Đặt B        = 1 x 2 + 2 x 3 + 3 x 4 + ... + 98 x 99

Suy ra 3B = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + ... + 98 x 99 x 100 - 97 x 98 x 99

                 = 98 x 99 x 100

B               = 98 x (99 : 3) x 100

B               = 98 x      33   x 100

Thay vào A được:

A = \(\frac{100x4851}{33x98x100}\) - 1

A =          \(\frac{3}{2}\)           - 1

A =          \(\frac{3}{2}\)           - \(\frac{2}{2}\)

A =                            \(\frac{1}{2}\)

Vậy A bằng \(\frac{1}{2}\)

Đáp số: \(\frac{1}{2}\)

=1x2x3+2x3x3+...+98x99x3

=1x2x3+2x3x(4-1)+...+98x99x(100-97)

=1x2x3+2x3x4-1x2x3+...+98x99x100-97x98x99

=98x99x100

=970200

330 nha

ta có :\(\frac{1}{1\cdot2}=\frac{1}{1}-\frac{1}{2}\) 

          \(\frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3}\)

           \(\frac{1}{3\cdot4}=\frac{1}{3}-\frac{1}{4}\)

            ......

          \(\frac{1}{99\cdot100}=\frac{1}{99}-\frac{1}{100}\)

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

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