Gọi biểu thức trên là A, ta có :

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

sai thì thui nhá,mk hok ngu lắm

ahihi

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{99\times100}\)

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

\(\frac{1}{1}-\frac{1}{100}\)

\(\frac{100-1}{100}\)

\(\frac{99}{100}\)

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{1}-\frac{1}{100}\)
\(\frac{100-1}{100}\)
\(\frac{99}{100}\)

 

99/100

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

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

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

            ......

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

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

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

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

S=1x2+2x3+3x4+4x5+...+98x99

3S= 1.2.3+ 2.3.3 + 3.4.3 + 4.5.3+...+98.99.3

3S= 1.2.3+ 2.3(4-1) + 3.4(5-2) + 4.5(6-3)+....+ 98.99.(100-97)

3S= 1.2.3 + 2.3.4 -1.2.3 + 3.4.5 - 2.3.4 +...+98.99.100 -97.98.99

3S= 98.99.100

S=970200:3

S= 323400

Bài làm:

\(S=1.2+2.3+3.4+...+98.99\)

\(S=\frac{1}{3}\left(1.2.3+2.3.3+3.4.3+...+98.99.3\right)\)

\(S=\frac{1}{3}\left[1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+98.99.\left(100-97\right)\right]\)

\(S=\frac{1}{3}\left(1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-97.98.99+98.99.100\right)\)

\(S=\frac{98.99.100}{3}=323400\)

Vậy S = 323400

Học tốt!!!!

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

1 \(\times\) 2 \(\times\) 3 = 1  \(\times\) 2 \(\times\) 3

2 \(\times\) 3 \(\times\) 3 = 2 \(\times\) 3 \(\times\) ( 4 -1) = 2  \(\times\) 3 \(\times\) 4 - 1 \(\times\) 2 \(\times\) 3

3 \(\times\) 4 \(\times\) 3 = 3 \(\times\) 4 \(\times\) ( 5 -2) = 3 \(\times\) 4 \(\times\) 5 - 2 \(\times\) 3 \(\times\) 4

4 \(\times\) 5 \(\times\) 3 = 4 \(\times\) 5 \(\times\) ( 6- 3) = 4 \(\times\) 5 \(\times\) 6 - 3 \(\times\) 4 \(\times\) 5 

..................................................................................

99\(\times\)100\(\times\)3 = 99\(\times\)100\(\times\)(101-98) =99\(\times\)100\(\times\)101 - 98\(\times\)99\(\times\)100

Cộng vế với vế ta được:

1\(\times\)2\(\times\)3 + 2\(\times\)3\(\times\)3 + 3\(\times\)4\(\times\)3+ ...+99\(\times\)100\(\times\)3 = 99\(\times\)100\(\times\)101

(1\(\times\)2 + 2\(\times\)3 + 3\(\times\)4 +...+99\(\times\)100)\(\times\)3 = 99\(\times\)100\(\times\)101

1\(\times\)2 + 2\(\times\)3 + 3\(\times\)4+...+99\(\times\)100 = (99 \(\times\)100 \(\times\)101):3

1\(\times\)2 + 2\(\times\)3 + 3\(\times\)4+...+99\(\times\)100 = 333 300

cảm ơn cô

Cho tổng trên là A

Ta co : 

\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)

\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=9\left(\frac{1}{1}-\frac{1}{100}\right)\)

\(A=9.\frac{99}{100}\)

\(A=\frac{891}{100}\)

1/1.2 +1/2.3 +1/3.4 +...+1/98.99 +1/99.100

=1-1/2+1/2-1/3+1/3-1/4+...+1/98-1/99+1/99-1/100

=1-1/100=100/100-1/100=99/100

Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

   \(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)

   \(\Rightarrow1-\frac{1}{100}=\frac{99}{100}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

= \(1-\frac{1}{100}=\frac{99}{100}\)