Đặt \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)

Ta có: \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)

\(\Leftrightarrow2A=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\dfrac{2}{3\cdot4\cdot5}+...+\dfrac{2}{98\cdot99\cdot100}\)

\(\Leftrightarrow2A=-\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}-\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}-\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}-\dfrac{1}{4\cdot5}+...-\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)

\(\Leftrightarrow2A=-\dfrac{1}{2}+\dfrac{1}{99\cdot100}\)

\(\Leftrightarrow2A=\dfrac{-1}{2}+\dfrac{1}{9900}\)

\(\Leftrightarrow2A=\dfrac{-4950}{9900}+\dfrac{1}{9900}=\dfrac{-4949}{9900}\)

hay \(A=\dfrac{-4949}{19800}\)

Câu b:

B = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ...+ 1/98.99.100

B = 1/2. (2/1.2.3 + 2/2.3.4 + ...+ 2/98.99.100)

B = 1/2.(1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ...+ 1/98.99 - 1/99.100)

B = 1/2.(1/2 - 1/9900)

B = 1/2.4949/9900

B = 4949/19800

\(C=1.2.3+2.3.4+........+48.49.50\)

\(\Rightarrow4C=1.2.3.4+2.3.4.4+........+48.49.50.4\)

\(=1.2.3.4+2.3.4.\left(5-1\right)+.........+48.49.50.\left(51-47\right)\)

\(=1.2.3.4+2.3.4.5-1.2.3.4+........+48.49.50.51-47.48.49.50\)

\(=48.49.50.51\)

\(\Rightarrow C=\frac{48.49.50.51}{4}=1499400\)

Ta có C = 1 x 2 x 3 + 2 x 3 x 4 + ... + 48 x 49 x 50

=>   4C  = 1 x 2 x 3 x 4 + 2 x 3 x 4 x 4 + ....  + 48 x 49 x 50 x 4

       4C   = 1 x 2 x 3 x 4 +  2 x 3 x 4 x (5 - 1)+ ... + 48 x 49 x 50 x (51 - 47)

       4C   = 1 x 2 x 3 x 4 + 2 x 3 x 4 x 5 - 1 x 2 x 3 x 4 + .... + 48 x 49 x 50 x 51 -  47 x 48 x 49 x 50 

       4C   = 48 x 49 x 50 x 51

       4C   = 5997600

         C   = 5997600 : 4

         C   = 1499400

Vậy C   = 1499400

Đặt A = 1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 +....+ 98 x 99 x 100

4A = 1 x 2 x 3 x 4 + 2 x 3 x 4 x 4 + 4 x 5 x 4 +....+ 98 x 99 x 100 x 4

4A = 1 x 2 x 3 x ( 4 - 0 ) + 2 x 3 x 4 x ( 5 - 1 ) + 4 x 5 x 6 x ( 7 - 3 ) +....+ 98 x 99 x 100 x ( 101 - 97 )

4A = 1 x 2 x 3 x 4 + 2 x 3 x 4 x 5 - 1 x 2 x 3 x 4 + 4 x 5 x 6 x 7 - 3 x 4 x 5 x 6 + .... + 98 x 99 x 100 x 101 - 98 x 99 x 100 x 97

A = 98 x 99 x 100 x 97 / 4

A = 98 x 99 x 25 x 97

ta có:
4s=1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+.........+k(k+1)(k+2)((k+3)-(k-1))
4s=1.2.3.4-1.2.3.0+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+........+k(k+1)(k+2)(k+3)-(k-1)k(k+1)(k+2)
4s=k(k+1)(k+2)(k+3)
ta biết rằng tích 4 số tự nhiên liên tiếp khi cộng thêm 1 luôn là 1 số chính phương
=>4s+1 là 1 số chính phương

A = 1.3 + 3.5 + ..+ 99.101

6A = 1.3.6 + 3.5.6 + 5.7.6 + ...+99.101.6

1.3.6 = 1.3.(5 + 1) = 1.3.5 + 1.3.1

3.5.6 = 3.5.(7 - 1) = 3.5.7 - 1.3.5

5.7.6 = 5.7.(9- 3) = 5.7.9 - 3.5.7

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

99.101.6 = 99.101.(103 - 97) = 99.101.103-97.99.101

Cộng vế với vế ta có:

6A = 1.3.1 + 99.101.103

6A = 3 + 9999.103

6A = 3 + 1029897

6A = 1029900

A = 1029900 : 6

A = 171650

Câu b:

B = 1.2.3 + 2.3.4 + 3.4.5 + ...+99.100.101

4B = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 +..+ 99.100.101.4

1.2.3.4 = 1.2.3.4

2.3.4.4 = 2.3.4.(5-1) = 2.3.4.5 - 1.2.3.4

3.4.5.4 = 3.4.5.(6 - 2) = 3.4.5.6 - 2.3.4.5

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

99.100.101.4 = 99.100.101.(102 - 98) = 99.100.101.102 - 98.99.100.101

Cộng vế với vế ta có:

4B = 99.100.101.102

B = 99.100.101.102 : 4

B = 25497450