\(1.4.7+4.7.10+...+n\left(n+3\right)\left(n+6\right)\\ =\dfrac{n^2\left(n+1\right)^2}{4}+9\cdot\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}+18\cdot\dfrac{n\left(n+1\right)}{2}\)

\(=\dfrac{n\left(n+1\right)\left(n^2+13n+42\right)}{4}=\dfrac{n\left(n+1\right)\left(n+6\right)\left(n+7\right)}{4}\)

Áp dụng vào bài toán:

\(P=\dfrac{2021.2022.2027.2028}{4}=...\)

CM: 

Với \(n=1\Leftrightarrow1.4.7=28\)

\(\dfrac{n\left(n+1\right)\left(n+6\right)\left(n+7\right)}{4}=\dfrac{2.7.8}{4}=28\)

Giả sử \(n=k\Leftrightarrow1.4.7+4.7.10+...+k\left(k+3\right)\left(k+6\right)=\dfrac{k\left(k+1\right)\left(k+6\right)\left(k+7\right)}{4}\)

Với \(n=k+1\), cần cm:

\(1.4.7+4.7.10+...+k\left(k+3\right)\left(k+6\right)+\left(k+1\right)\left(k+4\right)\left(k+7\right)=\dfrac{\left(k+1\right)\left(k+2\right)\left(k+7\right)\left(k+8\right)}{4}\)

Ta có \(VT=\dfrac{k\left(k+1\right)\left(k+6\right)\left(k+7\right)}{4}+\left(k+1\right)\left(k+4\right)\left(k+7\right)\)

\(=\left(k+1\right)\left(k+7\right)\left[\dfrac{k\left(k+6\right)}{4}+k+4\right]=\left(k+1\right)\left(k+7\right)\left(\dfrac{k^2+10k+16}{4}\right)\\ =\dfrac{\left(k+1\right)\left(k+7\right)\left(k+2\right)\left(k+8\right)}{4}=VP\)

Do đó theo pp quy nạp ta đc đpcm

=2.(6/1.4.7 + 6/4.7.10 + 6/7.10.13 + ... + 6/54.57.60)

=2.(1/1.4-1/4.7+1/4.7-1/7.10+1/7.10-1/10.13+...+1/54.57-1/57.60)

=2(1.4-1/57.60)

TỰ TÍNH

Ta có \(A=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+\frac{1}{7.10}-...+\frac{1}{16.19}-\frac{1}{19.22}\)

\(=\frac{1}{4}-\frac{1}{418}=\frac{207}{836}\)

\(A=\frac{6}{1\cdot4\cdot7}+\frac{6}{4\cdot7\cdot10}+\frac{6}{7\cdot10\cdot13}+...+\frac{6}{16\cdot19\cdot22}\)

\(A=\frac{1}{1\cdot4}-\frac{1}{4\cdot7}+\frac{1}{4\cdot7}-\frac{1}{7\cdot10}+...+\frac{1}{16\cdot19}-\frac{1}{19\cdot22}\)

\(A=\frac{1}{4}-\frac{1}{19\cdot22}=\frac{207}{836}\)

\(\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}\)

\(=3.\left(\frac{1}{1.4}-\frac{1}{4.7}\right)+3.\left(\frac{1}{4.7}-\frac{1}{7.10}\right)+...+3.\left(\frac{1}{54.57}-\frac{1}{57.60}\right)\)

\(=3\left(\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...-\frac{1}{54.57}+\frac{1}{54.57}-\frac{1}{57.60}\right)\)

\(=3\left(\frac{1}{1.4}-\frac{1}{57.60}\right)\)

\(=3\left(\frac{1}{4}-\frac{1}{3420}\right)\)

\(=3\left(\frac{855}{3420}-\frac{1}{3420}\right)\)

\(=3.\frac{427}{1710}\)

\(=\frac{427}{570}\)

\(A=\frac{1}{1.4.7}+\frac{1}{4.7.10}+...+\frac{1}{54.57.60}\)

\(\Rightarrow6A=\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}\)

\(=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{54.47}-\frac{1}{57.60}\)

\(=\frac{1}{4}-\frac{1}{3420}=\frac{855}{3420}-\frac{1}{3420}=\frac{427}{1710}\)

\(\Rightarrow A=\frac{427}{1710}:6=\frac{427}{1710}.\frac{1}{6}=\frac{427}{10260}\)

Nhận thấy: 

\(\frac{6}{1.4.7}=\frac{1}{1.4}-\frac{1}{4.7}\)

...............

\(\frac{6}{54.57.60}=\frac{1}{54.57}-\frac{1}{57.60}\)

=> ta phải nhân A vói 6 

=> 6A = 

\(\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{54.57}-\frac{1}{57.60}=\frac{1}{4}-\frac{1}{57.60}=\frac{427}{1710}\)

=> A = 427/1710 : 6 =427/10260

Đặt \(\frac{12}{1.4.7}+\frac{12}{4.7.10}+...+\frac{12}{54.57.60}=A\)

\(\frac{A}{2}=\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}\)

\(\frac{A}{2}=\frac{7-1}{1.4.7}+\frac{10-4}{4.7.10}+...+\frac{60-54}{54.57.60}\)

\(\frac{A}{2}=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{54.57}-\frac{1}{57.60}=\frac{1}{1.4}-\frac{1}{57.60}\)

\(A=\frac{1}{2}-\frac{1}{30.57}< \frac{1}{2}\)