Tinh b 1.4 2.5 2023.2023

a) 1 + (-2) + 3 + (-4) + .. + 19 + (-20)

= (-1) + (-1) + ... + (-1)     (có 10 số -1)

= (-1) . 10

= -10

b) 1 - 2 + 3 - 4 + ... + 99 - 100

= (-1) + (-1) + ... + (-1)   (có 50 số -1)

= (-1) . 50

= -50

c) 2 - 4 + 6 - 8 + ... + 48 - 50

= (-2) + (-2) + ... + (-2)   (có 25 số -2)

= (-2) . 25

= -50

d) -1 + 3 - 5 + 7 - ... + 97 - 99

= (-1) + (-2) + (-2) + ... (-2)   (có 49 số -2)

= (-1) + (-2) . 49

= (-1) + (-98)

= -99

e) 1 + 2 - 3 - 4 + ... + 97 + 98 - 99 - 100

= 1 + 2 - 3 - 4 + ... + 97 + 98 - 99 - 100 + 101 (ta cộng thêm 101 cho dễ tính)

= 1 + (2 - 3 - 4 + 5) + ... + (98 - 99 - 100 + 101)

= 1 + 0 + ... + 0

= 1 - 101 (ta bớt 101 để ra kết quả vì lúc nãy thêm 101)

= -100

\(A=1\left(2+2\right)+2\left(2+3\right)+3\left(2+4\right)+.....+\left(n-1\right)\left(2+n\right)\)

\(\Leftrightarrow A=1.2+1.2+2.3+2.2+3.4+2.3+....+\left(n-1\right)n+2\left(n-1\right)\)

\(\Leftrightarrow A=\left(1.2+2.3+.....+\left(n-1\right)n\right)+2\left(1+2+3+....+\left(n-1\right)\right)\)

Giả sử A=B+C

Với \(\begin{cases}B=1.2+2.3+.....+\left(n-1\right)n\\C=2\left[1+2+....+\left(n-1\right)\right]\end{cases}\)

Ta có

\(3B=1.2.\left(3-0\right)+2.3.\left(4-1\right)+......+\left(n-1\right)n\left[\left(n+1\right)-\left(n-2\right)\right]\)

\(\Rightarrow3B=1.2.3-0.1.2+2.3.4-1.2.3+.....+\left(n-1\right)n\left(n+1\right)-\left(n-2\right)\left(n-1\right)n\)

\(\Rightarrow B=\frac{\left(n-1\right)n\left(n+1\right)}{3}\)

Mặt khác

\(C=2\left[1+2+....+\left(n-1\right)\right]\)

\(\Rightarrow C=2.\frac{\left[\left(n-1\right)+1\right]n}{2}=n^2\)

\(\Rightarrow A=\frac{\left(n-1\right)n\left(n+1\right)}{3}+n^2\)

Vậy \(A=\frac{\left(n-1\right)n\left(n+1\right)}{3}+n^2\)

D =1.4+2.5+3.6+.......+99.102

D = 1. (2+2) +2.(2+3) +3.(2+4)+...+99.(100+2)

D = 1.2+1.2+2.2+2.3+2.3+3.4+...+2.99+99.100

D = (1.2+2.3+3.4+...+99.100) +2.(1+2+3+4+...+99)

*Gọi A= 1.2+2.3+3.4+...+99.100

3A = 3.(1.2+2.3+3.4+...+99.100)

3A = 1.2.3+2.3.3+...+99.100.3

3A = 1.2.3 +2 .3.(4-1)+...+99.100.(101-98)

3A = 1.2.3+2.3.4-1.2.3+...+ 99.100.101-98.99.100

3A = 99.100.101

3A = 3.33.100.101

A   =  33.100.101

A   = 333300

* Gọi B = 2. (1+2+3+4+...+99) 

                  \__có 99 số hạng ___/          

 B=  2.[(1+99).99:2]

 B = 2 .4950

 B = 9900

A+B = 333300+9900 =343200

Vậy D =343200

Đặt A = 1.4 + 2.5 + 3.6 + ... + 100.103

= 1.(2 + 2) + 2.(3 + 2) + 3.(4 + 2) +.... + 100.(101 + 2)

= 1.2 + 2.3 + 3.4 + ... + 100.101 + (1.2 + 2.2 + 3.2 + ... + 100.2)

= 1.2 + 2.3 + 3.4 + ... + 100.101 + 2(1 + 2 + 3 + .... + 100)

= 1.2 + 2.3 + 3.4 + .... + 100.101 + 2.100.(100 + 1) : 2

= 1.2 + 2.3 + 3.4 + ... + 100.101 + 10100

Đặt B = 1.2 + 2.3 + 3.4 + .... + 100.101

=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + .... + 100.101.3

=> 3B = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 100.101.(102 - 99)

=> 3B = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 100.101.102 - 99.100.101

=> 3B = 100.101.102

=> B = 343400

Khi đó A = 343400 - 10100 = 333300