1.Gộp 3 số vào thành 1 tổng rồi tính:

(1+2^1+2^2)+(2^3+2^4+2^5)+....+(2^37+2^38+2^39)

=1*(1+2^1+2^2)+2^3*(1+2^1+2^2)+....+2^37*(1+2^1+2^2)

=1*15+2^3*15+...+2^37*15

=15*(1+2^3+...+2^39) chia hết cho 15

\(1^n+2^n+3^n+4^n\)

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

\(=\left(4+1\right)\left(4^{n-1}-4^{n-2}+...-4+1\right)+\left(2+3\right)\left(2^{n-1}-2^{n-2}.3+...-2.3^{n-2}+3^{n-1}\right)\)

\(=5\left(4^{n-1}-4^{n-2}+...-4+1\right)+5\left(2^{n-1}-2^{n-2}.3+...-2.3^{n-2}+3^{n-1}\right)⋮5\)(đpcm)

Vậy \(1^n+2^n+3^n+4^n⋮5\)

c) \(n\left(2n-3\right)-2n\left(n+1\right)\)

\(=2n^2-3n-2n^2-2n\)

\(=-5n\)Vì n nguyên

\(\Rightarrow-5n⋮5\left(đpcm\right)\)

a) \(\left(2n+3\right)^2-9\)

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

\(=2n\left(2n+6\right)\)

\(=4n\left(n+3\right)\)

Do \(n\in Z\Rightarrow n+3\in Z\)

\(\Rightarrow4n\left(n+3\right)⋮4\left(đpcm\right)\)

b) \(n^2\left(n+1\right)+2n\left(n+1\right)\)

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

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

Vì \(n\in Z\Rightarrow\left\{{}\begin{matrix}x+1\in Z\\n+2\in Z\end{matrix}\right.\)

Mà n,n+1,n+2 là 3 sô nguyên liên tiếp

\(\Rightarrow n\left(n+1\right)\left(n+3\right)⋮6\left(dpcm\right)\)

Ai có lời giải k ạ

\(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)

\(=6n^2+30n+n+5-\left(6n^2-3n+10n-5\right)\)

\(=6n^2+31n+5-6n^2-7n+5\)

\(=24n+10\)

\(=2\left(12n+5\right)\) chia hết cho 2

=> \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)chia hết cho 2 (Đpcm)