A=1.2.3+2.3.4+...+n.(n+1).(n+2)

=>4A=1.2.3.4+2.3.4.4+n(n+1)(n+2).4

=1.2.3.(4-0)+2.3.4.(5-1)+...+n.(n+1)(n+2)[(n+3)-(n-1)]

=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+n(n+1)(n+2)(n+3)-(n-1)-n.(n+1).(n+2).(n+3)

=n.(n+1)(n+2)(n+3)

=>4A+1=n(n+1)(n+2)(n+3)+1

=n.(n+3).(n+1)(n+2)+1

=(n²+3n).[n.(n+2)+1.(n+2)]+1

=(n²+3n).(n²+2n+n+2)+1

=(n²+3n).(n²+3n+2)+1

Đặt y=n²+3n

=>4A+1=y.(y+2)+1

=y²+2y+1

=y²+y+y+1

=y.(y+1)+(y+1)

=(y+1)(y+1)

=(y+1)²

Vậy 4A+1 là số chính phương

\(N=1.2.3+2.3.4+...+n\left(n+1\right)\left(n+2\right)\)

\(4N=1.2.3.4+2.3.4.4+...+4n\left(n+1\right)\left(n+2\right)\)

\(4N=1.2.3.4+2.3.4.\left(5-1\right)+....+n\left(n+1\right)\left(n+2\right)\left[\left(n+3\right)-\left(n-1\right)\right]\)

\(4N=1.2.3.4+2.3.4.5-1.2.3.4+...+n\left(n+1\right)\left(n+2\right)\left(n+3\right)-\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)

\(4N=n\left(n+1\right)\left(n+2\right)\left(n+3\right)\)

\(4N+1=n\left(n+1\right)\left(n+2\right)\left(n+3\right)+1\)

\(=n\left(n+3\right)\left(n+1\right)\left(n+2\right)+1\)

\(=\left(n^2+3n\right)\left(n^2+2n+n+2\right)+1\)

\(=\left(n^2+3n\right)\left(n^2+3n+2\right)+1\)

\(=\left(n^2+3n+1-1\right)\left(n^2+3n+1+1\right)+1\)

\(=\left(n^2+3n+1\right)^2-1+1=\left(n^2+3n+1\right)^2=t^2\)(1 số bất kì thỏa mãn)

Vậy \(4N+1\) là số chính phương (đpcm)

Ta có: \(E=1.2.3+2.3.4+.....+n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow4E=1.2.3.4+2.3.4.\left(5-1\right)+......+n\left(n+1\right)\left(n+2\right)\left[\left(n+3\right)-\left(n-1\right)\right]\)

\(\Rightarrow4E=1.2.3.4+2.3.4.5-1.2.3.4+....+\) \(n\left(n+1\right)\left(n+2\right)\left(n+3\right)-\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow4E=n\left(n+1\right)\left(n+2\right)\left(n+3\right)\)

\(\Rightarrow4E=n\left(n+3\right)\left(n+1\right)\left(n+2\right)=\left(n^2+3n\right)\left(n^2+3n+2\right)\)

Đặt n² + 3n +1 = y

\(\Rightarrow4E+1=\left(y-1\right)\left(y+1\right)+1=y^2-1+1=y^2\)

\(\Rightarrow4E+1=\left(n^2+3n+1\right)^2\)

Vì n tự nhiên => n² + 3n + 1 tự nhiên => 4E + 1 là số chính phương

=> đpcm.

nhanh hk

\(1a.\)

Ta có: \(n^4+4=\left(n^2\right)^2+4n^2+4-4n^2=\left(n^2+2\right)^2-\left(2n\right)^2=\left(n^2-2n+2\right)\left(n^2+2n+2\right)\)

Vì  \(n^2+2n+2>n^2-2n+2\)  với mọi  \(n\in N\) 

nên để  \(n^4+4\)  là số nguyên tố thì  \(n^2-2n+2=1\)  \(\Leftrightarrow\)  \(\left(n-1\right)^2=0\)  \(\Leftrightarrow\)  \(n-1=0\)  \(\Leftrightarrow\)  \(n=1\)

Vậy, với  \(n=1\)  thì   \(n^4+4\)  là số nguyên tố

bài 1a trong tờ 

c) \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

Vì: a-b+b-c+c-a=0

Sau đó xét các TH