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

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

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

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

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

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

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

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

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

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

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

...

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

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

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

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

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

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

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

hok tốt!

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

A = 1.2 + 2.3 + 3.4 + ... + 99.100

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

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

3A = 99.100.101

A = 33.100.101

A = 333300

\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{2014.2015}\)

\(\Leftrightarrow\frac{1}{4}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)

\(\Leftrightarrow\frac{1}{4}A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)

\(\Leftrightarrow\frac{1}{4}A=\frac{1}{1}-\frac{1}{2015}\)

\(\Leftrightarrow\frac{1}{4}A=\frac{2014}{2015}\)

\(\Leftrightarrow A=\frac{2014}{2015}\div\frac{1}{4}\)

\(\Leftrightarrow A=\frac{8056}{2015}\)

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