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

\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{6.7.8}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{6.7}-\frac{1}{7.8}\)

\(=\frac{1}{1.2}-\frac{1}{7.8}\)

\(=\frac{1}{2}-\frac{1}{56}\)

\(=\frac{28}{56}-\frac{1}{56}=\frac{27}{56}\)

Dấu . là nhân nha

\(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\)

\(\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\)

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

\(\frac{2}{6.7.8}=\frac{1}{6.7}-\frac{1}{7.8}\)

S= \(\frac{1}{1.2}-\frac{1}{7.8}=\frac{27}{56}\)

Lời giải:
$4S=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+...+k(k+1)(k+2)[(k+3)-(k-1)]$

$=[1.2.3.4+2.3.4.5+3.4.5.6+...k(k+1)(k+2)(k+3)]-[0.1.2.3+1.2.3.4+2.3.4.5+....+(k-1)k(k+1)(k+2)]$

$=k(k+1)(k+2)(k+3)$

$\Rightarrow 4S+1=k(k+1)(k+2)(k+3)+1=[k(k+3)][(k+1)(k+2)]+1$

$=(k^2+3k)(k^2+3k+2)+1=(k^2+3k)^2+2(k^2+3k)+1=(k^2+3k+1)^2$

$\Rightarrow 4S+1$ là số chính phương.

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

ta co:1/1*2*3=(1/1*2-1/2*3):2
1/2*3*4=(1/1*2-1/2*3):2
...
cu nhu the cho den:
1/98*99*100=(1/98*99-1/99*100):2
suy ra : 1/1*2*3+1/2*3*4+1/3*4*5+...+1/98*99*100
=(1/1*2-1/2*3):2+(1/2*3-1/3*4):2+...+(1/98*99-1/99*100):2
=(1/1*2-1/2*3+1/2*3-1/3*4+...+1/98*99-1/99*100):2
=(1/1*2-1/99*100):2
=(1/2-1/9900)
=(4950/9000-1/9000):2
=4949/9000:2
=4949/18000
học tốt

2/1×2×3 + 2/2×3×4 + 2/3×4×5 + ... + 2/36×37×38 + 2/37×38×39

= 1/1×2 - 1/2×3 + 1/2×3 - 1/3×4 + 1/3×4 - 1/4×5 + ... + 1/36×37 - 1/37×38 + 1/37×38 - 1/38×39

= 1/1×2 - 1/38×39

= 1/2 - 1/1482

= 370/741

\(\text{Ta có: }\) \(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}+....+\frac{1}{37.38}-\frac{1}{38.39}\)

                  \(=\frac{1}{2}-\frac{1}{38.39}\)

                  \(=\frac{1}{2}-\frac{1}{1482}\)

                  \(=\frac{370}{741}\)

\(n\left(n+1\right)\left(n+2\right)=\frac{1}{4}n\left(n+1\right)\left(n+2\right).4=\frac{1}{4}n\left(n+1\right)\left(n+2\right)\left[\left(n+3\right)-\left(n-1\right)\right]\)

\(=-\frac{1}{4}\left(n-1\right)n\left(n+1\right)\left(n+2\right)+\frac{1}{4}n\left(n+1\right)\left(n+2\right)\left(n+3\right)\)

\(4S=-0.1.2.3+1.2.3.4-1.2.3.4+2.3.4.5-....-\left(k-1\right)k\left(k+1\right)\left(k+2\right)+k\left(k+2\right)\left(k+2\right)\left(k+3\right)\)

\(=k\left(k+1\right)\left(k+2\right)\left(k+3\right)\)

\(4S+1=\left(k^2+3k\right)\left(k^2+3k+2\right)+1=\left(k^2+3k\right)^2+2.\left(k^2+3k\right)+1\)

\(=\left(k^2+3k+1\right)^2\)

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

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