a=1.2+2.3+3.4+4.+....+200.201

3A = 1.2.(3 - 0) + 2.3.(4 - 1) + .... + 200.201.(202 - 199)

3A = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + .... + 200.201.202

3A = 200.201 . 202

A = 2706800

\(A=1.2+2.3+3.4+...+200.201\)

\(\frac{1}{A}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{200.201}\)

\(\frac{1}{A}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{200}-\frac{1}{201}\)

\(\frac{1}{A}=\frac{1}{1}-\frac{1}{201}=\frac{200}{201}\)

\(A=1:\frac{200}{201}=\frac{1.201}{200}=\frac{201}{200}\)

( Số cuối + Số đầu ) . số số hạng : 2 =a

có lỗi ko

Ta có : A = 1.2 + 2.3 + 3.4 + ..... + 49.50

=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + 49.50.51

=> 3A = 49.50.51

= >A = 49.50.51/3 = 41650

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

3A= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +...+-99.100.98

3A=99.100.101

A= 33.100.101

Ta có : A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)

\(\Rightarrow\)3A = 1.2.(3-0)+2.3.(4-1)+3.4.(5-2).....n.(n+1).[(n+2)-(n-1)]

\(\Rightarrow\)3A= 1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+....+n.(n+1)(n+2)-(n-1)n(n+1)

\(\Rightarrow\)3A= (1.2.3-1.2.3)+(2.3.4-2.3.4)+....+[(n-1).n.(n+1)-(n-1)n(n+1)]+n.(n+1)(n+2)

\(\Rightarrow\)3A=n.(n+1)(n+2)

\(\Rightarrow\)A=\(\frac{\text{n.(n+1)(n+2)}}{3}\)

Tại sao có 3A

cách mình đúng;

3S = 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 + 2.3.4 - 2.3 + 3.4.5 - 2.3.4 + ... + n(n + 1)(n + 2) - n(n + 1)(n - 1)
= n(n + 1)(n + 2)
=> S = n(n + 1)(n + 2)/3

3A=1.2.3+2.3.3+3.4.3+...+19.20.3

3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+19.20.(21-18)

3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+19.20.21-18.19.20

3A=19.20.21

=> \(A=\frac{19.20.21}{3}=2660\)

mk dùng cách của lớp 8 nha bạn ;

ta có công thức xích ma như sau x(x+1)

nhập vào xích ma ta có kết quả 2660

Ta có;

A=1.2+2.3+3.4+.....+19.20

3A=1.2.3+2.3.(4-1)+3.4.(5-2)+....+19.20.(21-18)

3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.....+19.20.21-18.19.20

A=19.20.21 = 2660

        3

Ta có: 3A=1.2.3+2.3.3+3.4.3+.....+99.100.3

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

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

3A=99.100.101

A=\(\frac{99.100.101}{3}\)

A=333300

Đặt A= 1.2+2.3 +.......+99.100

3A= 1.2.3+2.3.4+3.4.3 +......+ 99.100.3

3A= 1.2. (3 - 0) + 2.3.(4 - 1) +3.4. (5 - 2)....... . 99.100. (101 - 98)

3A = (1.2.3 + 2.3.4 + 3.4.5 +...... + 99.100.101) - (0.1.2 + 1.2.3 + 2.3.4 +.......+ 98.99.100)

3A = 99.100.101 - 0.1.2

3A = 999900 - 0

3A= 999900

A= 999900 : 3

A = 333300

3A=1×2×3+2×3×3+3×4×3+...+99×100×3

3A=1×2×(3-0)+2×3×(4-1)+3×4×(5-2)+...+99×100×(101-98)

3A=1×2×3 + 2×3×4 - 2×3×1 + 3×4×5 - 3×4×2 +...+ 99×100×101 - 99×100×98

3A=99×100×101

=>A=99×100×101/3=333300

Tk mình nha bn !

\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{2019.2020}\)

\(=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)

\(=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)

\(=9\left(1-\frac{1}{2020}\right)\)

\(=9.\frac{2019}{2020}\)

\(=\frac{18171}{2020}\)

\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{2019.2020}\)

\(A=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)

\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)

\(A=9\left(1-\frac{1}{2020}\right)=\frac{9.2019}{2020}=\frac{18171}{2020}\)

...

\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{2019.2020}\)

\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)

\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)

\(A=9.\left(1-\frac{1}{2020}\right)\)

\(A=9.\left(\frac{2020}{2020}-\frac{1}{2020}\right)\)

\(A=9.\frac{2019}{2020}\)

\(A=\frac{18171}{2020}\)

hok tốt!