BN thử vào câu hỏi tương tự xem có k?

Nếu có thì bn xem nhé!

Nếu k thì xin lỗi đã làm phiền bn

Hội con 🐄 chúc bạn học tốt!!!

3ⁿ⁺² -2ⁿ⁺² +3ⁿ -2ⁿ

=3ⁿ .3² -2ⁿ .2² +3ⁿ -2²

=3ⁿ(9+)-2ⁿ(4-1)

Vì 3ⁿ .10 ⋮10

=> 3ⁿ .10- 2ⁿ .3⋮10

=>3ⁿ +2 -2ⁿ⁺² +3ⁿ -2ⁿ ⋮10

Bài 1:

$A=(n-1)(2n-3)-2n(n-3)-4n$

$=2n^2-5n+3-(2n^2-6n)-4n$

$=-3n+3=3(1-n)$ chia hết cho $3$ với mọi số nguyên $n$

Ta có đpcm.

Bài 2:
$B=(n+2)(2n-3)+n(2n-3)+n(n+10)$

$=(2n-3)(n+2+n)+n(n+10)$

$=(2n-3)(2n+2)+n(n+10)=4n^2-2n-6+n^2+10n$

$=5n^2+8n-6=5n(n+3)-7(n+3)+15$

$=(n+3)(5n-7)+15$

Để $B\vdots n+3$ thì $(n+3)(5n-7)+15\vdots n+3$

$\Leftrightarrow 15\vdots n+3$
$\Leftrightarrow n+3\in\left\{\pm 1;\pm 3;\pm 5;\pm 15\right\}$

$\Rightarrow n\in\left\{-2;-4;0;-6;-8; 2;12;-18\right\}$

Đây nè bạn

=>(3^n+2)+(3^n)-(2^n+2)-(2^n)=3^n((3^2)+1)-2^n((2^2)+1)=(3^n)*10-(2^n)*5=(3^n)*10-(2^n-1)*5*2=(3^n)*10-(2^n-1)*10=10*((3^n)-(2^n-1) chia hết cho 10

=>(3^n+2)-(2^n+2)+(3^n)-(2^n)chia hết cho 10

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

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

\(=6n+6=6\left(n+1\right)⋮6\forall n\in Z\)

\(3^{n+2}-2^{n+2}+3^n-2^n\)

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

\(=3^n.10-2^n.5\)

\(=3^n.10-2^{n-1}.10\)

\(=10.\left(3^n-2^{n-1}\right)⋮10\) ∀n∈N

Vậy ...

Tham khảo

\(=3^n.9-2^n.4+3^n-2^n\)

\(=10.3^n-5.2^n\)

\(=10.\left(3^n-2^n\right)\)

\(\Leftrightarrow⋮10̸\)

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

=\(^{3^n}\).9 - \(2^n\).4 +\(^{3^n}\)- \(2^n\)

=10 .\(3^n\)-5.\(2^n\)

=10.\(3^n\)-5.2.\(2^{n-1}\)

=10 .(\(3^n\)-\(2^n\) )

=> chia hết cho 10

Ta có: \(3^{n+2}-2^{n+2}+3^n-2^n\)

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

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

\(=3^n\cdot10-2^n\cdot5\)

\(=3^n\cdot10-2^{n-1}\cdot2\cdot5\)

\(=3^n\cdot10-2^{n-1}\cdot10\)

\(=\left(3^n-2^{n-1}\right)\cdot10⋮10\left(dpcm\right)\)