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![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt S = 1x2+2x3+3x4+...+98x99+99x100
S x 3 =1x2x3+2x3x3+3x4x3+...+98x99x3+99x100x3
S x 3 =1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+....+98x99x(100-97)+99x100x(101-98)
S x 3 = 1x2x3 + 2x3x4-1x2x3+3x4x5-2x3x4+...+98x99x100-97x98x99+99x100x101-98x99x100
S x 3 = 99x100x101
S x 3 = 999900
S = 333300
![](https://rs.olm.vn/images/avt/0.png?1311)
Kết quả là 3333
♣ Còn phần công thức, bạn có thể xem ở link sau bài 3 : https://sites.google.com/site/toantieuhocpl/20-tinh-nhanh
Gọi biểu thức trên là S, ta có :
100S = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
100S x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
100S x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
100S x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
100S x 3 = 99x100x101
100S = 99x100x101 : 3
100S = 333300
S=333300:100
S=3333
![](https://rs.olm.vn/images/avt/0.png?1311)
Cho tổng trên là A
Ta co :
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
3M = 1.2.3 + 2.3.(4-1) +..+ 99.100.(101-98)
3M = 1.2.3 + 2.3.4 - 1.2.3 + .... + 99.100.101 - 98.99.100
3M = 99.100.101 = 999900
M = 333300
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+....+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9\cdot\frac{99}{100}=\frac{891}{100}\)
\(A=9\left(\frac{1}{1x2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
=> \(A=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
=> \(A=9\left(1-\frac{1}{100}\right)=\frac{9.99}{100}=\frac{891}{100}\)
=> A=8,91
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1.98+2.97+3.96+...+98.1}{1.2+2.3+3.4+...+98.99}=\frac{1.\left(100-2\right)+2\left(100-3\right)+3\left(100-4\right)+...+98\left(100-99\right)}{1.2+2.3+3.4+...+98.99}\)
\(A=\frac{1.100-1.2+2.100-2.3+3.100-3.4+...+98.100-98.99}{1.2+2.3+3.4+...+98.99}\)
\(A=\frac{\left(1.100+2.100+3.100+...+98.100\right)-\left(1.2+2.3+3.4+...+98.99\right)}{1.2+2.3+3.4+...+98.99}\)
\(A=\frac{100\left(1+2+3+...+98\right)}{1.2+2.3+3.4+...+98.99}-1\)
Ta có: 1+2+3+...+98=98.99:2=4851
Đặt B=1.2+2.3+3.4+...+98.99 => 3B=1.2.3+2.3.3+3.4.3+...+98.99.3 = 1.2.3+2.3.(4-1)+3.4(5-2)+...+98.99(100-97)
=> 3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+98.99.100-97.98.99 = 98.99.100
=> B=33.98.100. Thay vào A được:
\(A=\frac{100.4851}{33.98.100}-1=\frac{3}{2}-1=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101
A = 99x100x101 : 3
A = 333300
hok tốt
ta có :
S = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
3S = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
3S = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
3S = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
3S = 99x100x101
S = 99x100x101 : 3
S = 333300
=> 100S = 333300 . 100 = 33330000
làm bừa thui,ai tích mình mình tích lại
Số số hạng là :
Có số cặp là :
50 : 2 = 25 ( cặp )
Mỗi cặp có giá trị là :
99 - 97 = 2
Tổng dãy trên là :
25 x 2 = 50
Đáp số : 50