a) \(A=2.4+4.6+6.8+...+18.20\)

\(6A=2.4.6+4.6.\left(8-2\right)+6.8.\left(10-4\right)+...+18.20.\left(22-16\right)\)

\(6A=2.4.6+4.6.8-2.4.6+6.8.10-4.6.8+...+18.20.22-16.18.20\)

\(6A=18.20.22\)

\(A=\frac{18.20.22}{6}=\frac{7920}{6}=1320\)

d/ Đặt : A = 1.2 + 2.3 + 3.4 + ......... + 99.100

=> 3A = 1.2.(3 - 0) + 2.3.(4 - 1) + ..... + 99.100.(101 - 98)

=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 99.100.101

=> 3A = 99.100.101

=> A = 99.100.101 / 3

=> A = 333300 

a/

3A=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)=

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

=98.99.100=> A=98.33.100

b

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

=1.3.(5+1)+3.5.(7-1)+5.7.(9-3)+...+99.101.(103-97)=

=1.3+1.3.5-1.3.5+3.5.7-3.5.7+5.7.9-...-97.99.101+99.101.103=

=1.3+99.101.103=> (3+99.101.103):6

c/

9S=1.4.9+4.7.9+7.10.9+...+2017.2020.9=

=1.4.(7+2)+4.7.(10-1)+7.10.(13-4)+...+2017.2020.(2023-2014)=

=1.2.4+1.4.7-1.4.7+4.7.10--4.7.10+7.10.13-...-2014.2017.2020+2017.2020.2023=

=1.2.4+2017.2020.2023=> S=(2.4+2017.2020.2023):9

Dạng tổng quát: tính tổng các tích có quy luật: các thừa số của các tích lập thành dãy số cách đều. các thừa số đầu tiên của số hạng liền sau cũng chính là các thừa số sau cùng của số hạng liền trước thì ta nhân tổng với số k

Số k được tính theo quy luật \(k=\left(n+1\right)xd\)

            Trong đó: n: số thừa số của 1 số hạng

                            d: Khoảng cách giữa hai thừa số liền kề trong mỗi số hạng

Chúc em học tốt

\(A=\) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}\)

\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.50}\)

A= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

A = \(\frac{1}{1}-\frac{1}{51}=\frac{50}{51}\)

\(A=1\cdot2+2\cdot3+...+151\cdot152\)

\(=1\left(1+1\right)+2\left(1+2\right)+...+151\left(1+151\right)\)

\(=\left(1+2+3+...+151\right)+\left(1^2+2^2+...+151^2\right)\)

\(=\dfrac{151\left(151+1\right)}{2}+\dfrac{151\left(151+1\right)\left(2\cdot151+1\right)}{6}\)

\(=151\cdot76+\dfrac{151\cdot152\cdot303}{6}\)

\(=151\cdot76+151\cdot7676=1170552\)

\(C=2\cdot4+4\cdot6+...+2024\cdot2026\)

\(=2\cdot2\left(1\cdot2+2\cdot3+...+1012\cdot1013\right)\)

\(=4\left[1\left(1+1\right)+2\left(1+2\right)+...+1012\left(1+1012\right)\right]\)

\(=4\left[\left(1+2+...+1012\right)+\left(1^2+2^2+...+1012^2\right)\right]\)

\(=4\left[1012\cdot\dfrac{1013}{2}+\dfrac{1012\left(1012+1\right)\left(2\cdot1012+1\right)}{6}\right]\)

\(=4\left[506\cdot1013+345990150\right]\)

\(=1386010912\)

\(M=1^2+2^2+...+2024^2\)

\(=\dfrac{2024\left(2024+1\right)\cdot\left(2\cdot2024+1\right)}{6}\)

\(=2024\cdot2025\cdot\dfrac{4049}{6}\)

=2765871900

\(N=1^3+2^3+...+100^3\)

\(=\left(1+2+3+...+100\right)^2\)

\(=\left[\dfrac{100\left(100+1\right)}{2}\right]^2\)

\(=\left[50\cdot101\right]^2=5050^2\)

\(Q=1^3+2^3+...+2024^3\)

\(=\left(1+2+3+...+2024\right)^2\)

\(=\left[\dfrac{2024\left(2024+1\right)}{2}\right]^2\)

\(=\left[1012\left(2024+1\right)\right]^2\)

\(=2049300^2\)

I.\(B=9,8+8,7+7,6+...+2,1-1,2-2,3-3,4-...-8,9\)

\(B=\left(9,8-8,9\right)+\left(8,7-7,8\right)+\left(7,6-6,7\right)+...+\left(2,1-1,2\right)\)

\(B=0,9+0,9+0,9+...+0,9\) ( 8 số 0,9 )

\(B=7,2\)

II.

\(\left(a\right)\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{19\cdot20}\)

\(=2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{19\cdot20}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{19}-\frac{1}{20}\right)\)

\(=2\left(1-\frac{1}{20}\right)\)

\(=2\cdot\frac{19}{20}=\frac{19}{10}\)

\(\left(b\right)\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{17\cdot19}+\frac{4}{19\cdot21}\)

\(=2\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{17\cdot19}+\frac{2}{19\cdot21}\right)\)

\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{21}\right)\)

\(=2\left(1-\frac{1}{21}\right)\)

\(=2\cdot\frac{20}{21}=\frac{40}{21}\)

\(\left(c\right)\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{16\cdot18}+\frac{4}{18\cdot20}\)

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}=\frac{9}{10}\)

cảm ơn  bạn

b: 6B=2*4*6+4*6*6+6*8*6+...+46*48*6+48*50*6

=2*4*6-2*4*6+4*6*8-4*6*8+...-44*46*48+46*48*50-46*48*50+48*50*52

=48*50*52

=>B=20800

d: 9D=1*4*9+4*7*9+...+46*49*9

=1*4*2+1*4*7-1*4*7+1*7*10-1*7*10+...+46*49*52-46*49*43

=1*2*4+46*49*52

=117216

=>D=13024

a: 

B = 1.2+2.3+3.4+...+99.100

B=1.100

B=100

C=1.3+2.4+3.5+4.6+...+9.11

C=1.(2+1)+2.(3+1)+3.(4+1)+4.(5+1)+...+9.(10+1)

C=1.2+1+2.3+1+3.4+1+4.5+1+...+9.10+1

C=(1.2+2.3+3.3+4.5+...+9.10)+(1+1+1+1+..+1)

C=1.10+10

C=10+10

C=20

a) B = 1.2+2.3+3.4+..+99.100

=>3B=1.2.3+2.3.3+3.4.3+...+99.100.3

3B = 1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)

3B = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5-2.3.4+...+99.100.101-98.99.100

3B = (1.2.3+2.3.4+3.4.5+..+99.100.101) - (1.2.3+2.3.4+...+98.99.100)

3B = 99.100.101

\(B=\frac{99.100.101}{3}=333300\)

b) C = 1.3+2.4+3.5+4.6+...+9.11

C = (2-1).(2+1)+(3-1).(3+1) + (4-1).(4+1)+(5-1).(5+1)+...+(10-1).(10+1)

C = 2² - 1 + 3² - 1 + 4² - 1 + 5² - 1 +...+10² - 1

C = (2²+3²+4²+5²+...+10²) -(1+1+...+1) 

...

Ta có :

A = 1.2+2.3+3.4+4.5+...+100.101

B = 1.3+2.4+3.5+4.6+....+100.102

=> B - A = ( 1.2+2.3+3.4+4.5+...+100.101) - (1.3+2.4+3.5+4.6+...+100.102)

=> B - A = 1.2+2.3+3.4+4.5+...+100.101-1.3-2.4-3.5-4.6-....-100.102

=> B - A = 1.2+(2.3-1.3)+(3.4-2.4)+(4.5-3.5)+...+(100.101-99.101)-100.102

=> B - A = 2+3+4+5+...+101-10200

=> B - A = (2+101)+(3+100)+...+(51+52)-10200

=> B - A = 103+103+103+....+103-10200 ( 50 SỐ 103 )

=> B - A = 103.50-10200

=> B - A = 5150-10200

=> B - A = -5050