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=5(1-1/2+1/2-1/3+...+1/2023-1/2024)
=5*2023/2024
=10115/2024

Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 32.33
=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 32.33.34
=> 3S = 32.33.34
=> S = \(\frac{32.33.34}{3}=11968\)
Cho A=1/1.2 + 1/2.3 + + 1/ 3.4+...+1/49.50 ; B = 1.2+2.3+3.4+4.5+5.6+...+49.50
Tính 50 mủ 2 A – B/17


3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2014.2015.(2016-2013)
3C=2014.2015.2016
C=2014.2015.2016:3

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
\(B=1.2+2.3+3.4+...+49.50\)
\(3B=1.2.3+2.3.3+3.4.3+...+49.50.3\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)
\(=49.50.51\)
\(B=\frac{49.50.51}{3}=49.50.17\)
\(50^2.A-\frac{B}{17}=49.50-49.50=0\)

C=1*2+2*3+3*4+...+98*99
C=2+6+12+...+9702
C=2+9702
C=9704
vay C=9704
D=(1*99+2*99+3*99+...+99*99)-(1*2+2*3+3*4+...+98*99)
D=(99+198+297+...+9801)-(2+6+12+...+9702)
D=(99+9801)-(2+9702)
D=9900-9704
D=196
vay D=196
ai di qua dong tinh thi nho h cho minh nhe

Ta có : 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 = 99.100.101
=> 3B = 999900
=> B = 333300
Vậy B = 333300
Bài làm :
Ta có :
B= 1.2 + 2.3 + 3.4 + ...+ 99.100
=>3B = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3
<=>3B= 1.2.3+2.3(4-1)+3.4(5-2)+...+98.99(100-97)+99.100(101-98)
<=>3B= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-97.98.99+99.100.101-98.99.100
<=>3S = 99.100.101
<=> 3S = 999900
<=> B = 999900 : 3 = 333300
Vậy B = 333300

D = 1.2 + 2.3+ 3.4 +...+ 99.100
=>3D=1.2.3+2.3.3+3.4.3+...+99.100.3
=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+....+99.100.(101-98)
=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100
=99.100.101-0.1.2
=99.100.101
=999900
=>D=999900:3=333300
Dn = 1.2 + 2.3 + 3.4 +...+ n (n +1)
=>3Dn=1.2.3+2.3.3+3.4.3+...+n(n+1).3
=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+n.(n+1).[(n+2)-(n-1)]
=1.2.3-0.1.2+2.3.4-1.2.3+2.3.4-2.3.4+....+n(n+1)(n+2)-(n-1)n(n+1)
=n.(n+1).(n+2)-0.1.2
=n.(n+1)(n+2)
=>Dn=n.(n+1)(n+2):3
=>điều cần chứng minh

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=99.100.101
=> B = 99.100.101:3 = 333300
Đặt B = 1.2 + 2.3 + 3.4 + ... + 99.100
3B = 1.2.3 + 2.3.4 + 3.4.3 + ... + 99.100.3
3B = 1.2(3-0)+2.3(4-1)+3.4(5-2). .... . 99.100 ( 101 - 98 )
3B = (1.2.3 + 2.3.4 + 3.4.5 + .. + 99.100.101) - (0.1.2 + 1.2.3 + 2.3.4 + ... + 98.99.100)
3B = 99.100.101 - 0.1.2
3B = 999900 - 0
3B = 999900
B = 999900 : 3
B = 333300
\(b=1\cdot2+2\cdot3+3\cdot4+...+2023\cdot2024\\ 3b=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot3\cdot4+...+2023\cdot2024\cdot3\\ 3b=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+2023\cdot2024\cdot\left(2025-2022\right)\\ 3b=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+2023\cdot2024\cdot2025-2022\cdot2023\cdot2024\\ 3b=\left(1\cdot2\cdot3-1\cdot2\cdot3\right)+\left(2\cdot3\cdot4-2\cdot3\cdot4\right)+...+\left(2022\cdot2023\cdot2024-2022\cdot2023\cdot2024\right)+2023\cdot2024\cdot2025\\ 3b=8291467800\\ b=2763822600\)
B = 1.2 + 2.3 + ... + 2023.2024
B = \(\dfrac{1}{3}\).(1.2.3 + 2.3.3 + ... + 2023.2024.3)
B = \(\dfrac{1}{3}\).[1.2.3 + 2.3.(4-1) + ...+ 2023.2024.(2025-2022)]
B = \(\dfrac{1}{3}\).[1.2.3 + 2.3.4- 1.2.3 + ... + 2023.2024.2025 - 2022.2023.2024]
B = \(\dfrac{1}{3}\).2023.2024.2025
B = 2023.2024.(2025.\(\dfrac{1}{3}\))
B = 4094552.675
B = 2763822600