Ta có: \(3^{n+2}-2^{n+2}+3^n-2^n\)

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

\(=3^n\cdot10-2^{n-1}\cdot10⋮10\)

`3^(n+2)-2^(n+2)+3^n-2^n`

`=3^n .3^2 -2^n .2^2+3^n-2^n`

`= 3^n .(3^2+1)-2^n .(2^2+1)`

`=3^n .10 - 2^n . 5`

`=3^n .10 - 2^(n-1) .2.5`

`=3^n .10 -2^(n-1) .10 vdots 10 `

Ta có : 3^{n + 2} - 2^{n + 2} + 3ⁿ - 2ⁿ

<=>  3^{n + 2} - 2^{n + 2} + 3ⁿ - 2ⁿ = (3^{n + 2} + 3) - (2^{n + 2} + 2ⁿ)

=> 3^{n + 2} - 2^{n + 2} + 3ⁿ - 2ⁿ = 3ⁿ.(3² + 1) - 2^{n - 1}.(2³ + 2)

=> 3^{n + 2} - 2^{n + 2} + 3ⁿ - 2ⁿ = 3ⁿ . 10 - 2^{n - 1}.10 

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

Mà 3ⁿ - 2^{n - 1} E N*

Nên  3^{n + 2} - 2^{n + 2} + 3ⁿ - 2ⁿ chia hết cho 10 cới mọi n e N*

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

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

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

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

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

Vậy ...

Tham khảo

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

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

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

\(\Leftrightarrow⋮10̸\)

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

\(=3^n.10-2^n.5=3^n.10-2^{n-1}.10=\left(3^n-2^{n-1}\right).10\) chia hết cho 10(đpcm)

tick tớ nhé

\(S=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)=10.3^n-5.2^n=10.3^n-5.2.2^{n-1}=10\left(3^n-2^{n-1}\right)⋮10\) (đpcm)

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