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: \(\frac{1}{2}\)
